(i) (ii) Solution: (i) As the two chords \[AB\text{ }and\text{ }CD\] intersect each other at \[P\] We have \[AP\text{ }x\text{ }PB\text{ }=\text{ }CP\text{ }x\text{ }PD\] \[4.5\text{ }x\text{...
The ratio between the altitudes of two similar triangles is same as the ratio between their sides. So, The ratio between the areas of two similar triangles is same as the square of the ratio between...
The ratio between the altitudes of two similar triangles is same as the ratio between their sides. So, (i) The ratio between the medians of two similar triangles is same as the ratio between their...
Let’s consider two similar triangles as \[\blacktriangle ABC\text{ }\sim\text{ }\blacktriangle PQR\] So, \[Ar\left( \blacktriangle ABC \right)/\text{ }Ar\left( \blacktriangle PQR \right)\] \[=\text{...
Solution: (i) In \[\vartriangle BFD\text{ }and\text{ }\vartriangle BEC,\] \[\angle BFD\text{ }=\angle BEC\] [Corresponding angles] \[\angle FBD\text{ }=\angle EBC\] [Common] Hence, \[\vartriangle...
Solution: In \[\vartriangle ABC,\text{ }as\text{ }PR\text{ }||\text{ }BC\text{ }by\text{ }BPT\]we have \[AP/PB\text{ }=\text{ }AR/RC\] And, in \[\vartriangle PAR\text{ }and\text{ }\vartriangle...
Solution: As, \[\vartriangle ACD\text{ }\sim\text{ }\vartriangle BCA\] We have, \[Ar\left( \vartriangle ACD \right)/\text{ }Ar\left( \vartriangle BCA \right)\] \[=\text{ }A{{D}^{2}}/\text{...
Solution: (i) In \[\vartriangle ACD\text{ }and\text{ }\vartriangle BCA\] \[\angle DAC\text{ }=\angle ABC\] [Given] \[\angle ACD\text{ }=\angle BCA\][Common angles] Hence, \[\vartriangle ACD\text{...
Solution: (i) In \[\vartriangle PCD\text{ }and\text{ }\vartriangle PEF\] \[\angle CPD\text{ }=\angle EPF\][Vertically opposite angles] \[\angle DCE\text{ }=\angle FEP\][As DC || EF, alternate...
Solution: Given, \[DE\text{ }||\text{ }AC\] So, \[BE/EC\text{ }=\text{ }BD/DA\text{ }\left[ By\text{ }BPT \right]\] And, \[DC\text{ }||\text{ }AP\] So, \[BC/CP\text{ }=\text{ }BD/DA\text{ }\left[...
Solution: Let join \[AR\]such that it intersects \[BQ\]at point \[X.\] In \[\vartriangle ACR,\text{ }BX\text{ }||\text{ }CR\] By\[BPT\], we have \[AB/BC\text{ }=\text{ }AX/XR\text{ }\ldots \text{...
Solution: According to the given question, \[\angle AXY\text{ }=~\angle AYX\] So, \[AX\text{ }=\text{ }AY\][Sides opposite to equal angles are equal.] Also, from BPT we have \[BX/AX\text{ }=\text{...
Solution: According to the given question, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR\] And, \[AD\text{ }and\text{ }PM\]are the angle bisectors. So, \[\angle BAD\text{ }=\angle QPM\]...
Solution: According to the given question, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR\] So, \[\angle ABC\text{ }=\angle PQR\] i.e. \[\angle ABD\text{ }=\angle PQM\] Also, \[\angle...
Solution: According to the given question, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR\] \[AD\text{ }and\text{ }PM\]are the medians, so \[BD\text{ }=\text{ }DC\text{ }and\text{ }QM\text{...
Solution: In \[\Delta \text{ }FDC\text{ }and\text{ }\Delta \text{ }FBA,\] \[\angle FDC\text{ }=\angle FBA\text{ }\left[ As\text{ }DC\text{ }||\text{ }AB \right]\] \[\angle DFC\text{ }=\angle BFA\]...
Solution: We already have, \[\vartriangle AEB\text{ }\sim\text{ }\vartriangle FEC\] So, \[AE/FE\text{ }=\text{ }BE/CE\] \[=\text{ }AB/FC\] \[AE/FE\text{ }=\text{ }9/13.5\] Or, \[\left( AF\text{...
Solution: (i) In \[\Delta \text{ }AEB\text{ }and\text{ }\Delta \text{ }FEC,\] \[\angle AEB\text{ }=\angle FEC\] [Vertically opposite angles] \[\angle BAE\text{ }=\angle CFE\] [Since, AB||DC] Hence,...
Solution: According to the given question, \[XY\text{ }||\text{ }BC\] So, In \[\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC\] \[\angle AXY\text{ }=\angle ABC\] [Corresponding angles]...
Solution: According to the given question, \[XY\text{ }||\text{ }BC\] So, In \[\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC\] \[\angle AXY\text{ }=\angle ABC\][Corresponding angles]...
(i)\[OA\text{ }=\text{ }6\text{ }cm\] So, \[OA\text{ }\left( 3 \right)\text{ }=\text{ }OA\] \[OA\text{ }\left( 3 \right)\text{ }=\text{ }6\] Or, \[OA\text{ }=\text{ }2\text{ }cm\] (ii) \[OC\text{...
According to the given question, \[\Delta \text{ }ABC\]is enlarged and the scale factor \[m\text{ }=\text{ }3\]to the \[\Delta \text{ }ABC\] (i) \[AB\text{ }=\text{ }4\text{ }cm\] So, \[AB\left( 3...
According to the given question, \[\Delta \text{ }LMN\] has been reduced by a scale factor \[m\text{ }=\text{ }0.8\text{ }to\text{ }\Delta \text{ }LMN\] (i) \[MN\text{ }=\text{ }8\text{ }cm\] So,...
Given that, \[\Delta \text{ }ABC\]has been enlarged by scale factor \[m\text{ }of\text{ }2.5\text{ }to\text{ }\Delta \text{ }ABC\] (i) \[AB\text{ }=\text{ }6\text{ }cm\] So, \[AB\left( 2.5...
Solution: Because, \[\Delta \text{ }LQM\text{ }and\text{ }\Delta \text{ }LQN\] have common vertex at \[L\]and their bases \[QM\text{ }and\text{ }QN\] are along the same straight line. \[Area\text{...
Solution: (i) In \[\Delta \text{ }PLM\text{ }and\text{ }\Delta \text{ }PQR\] As LM || QR, corresponding angles are equal. \[\angle PLM\text{ }=\angle PQR\] \[\angle PML\text{ }=\angle PRQ\] So,...
It’s given that, \[Ar\left( \Delta \text{ }APQ \right)\]\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }Ar\left( \Delta \text{ }ABC \right)\] \[Ar\left( \Delta \text{ }APQ \right)/\text{...
Solution: According to the given question, \[AX/XB\text{ }=\text{ }3/5\] \[\Rightarrow AX/AB\text{ }=\text{ }3/8\text{ }\ldots .\text{ }\left( 1 \right)\] (i) In \[\Delta \text{ }AXY\text{...
Suppose, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle DEF\] So, \[AB/DE\text{ }=\text{ }BC/EF\] \[=\text{ }AC/DF\] Or, \[=\text{ }\left( AB+BC+AC \right)/\left( DE+EF+DF \right)\] \[=\text{...
According to the given question, \[AP\text{ }=\text{ }\left( 1/3 \right)\text{ }PB\] So, \[AP/PB\text{ }=\text{ }1/3\] In \[\vartriangle \text{ }APQ\text{ }and\text{ }\vartriangle ABC\] As\[PQ\text{...
As per the given question, The ratio of the areas of two similar triangle are equal to the ratio of squares of their corresponding sides. Thus, (i) The ration is, (ii) The ratio is,
Solution: According to the given question, \[\Delta \text{ }ABC\text{ }\sim\text{ }\Delta \text{ }ADE\] So, we have \[AE/AC\text{ }=\text{ }DE/BC\] \[4/11\text{ }=\text{ }6.6/BC\] Or, \[BC=\left(...
(i) In \[\vartriangle \text{ }ADE\text{ }and\text{ }\vartriangle \text{ }ABC\] \[AE/EC\text{ }=\text{ }6/7.5\text{ }=\text{ }4/5\] \[AD/BD\text{ }=\text{ }4/5\] \[\left[ BD\text{ }=\text{ }AB\text{...
In \[\vartriangle \text{ }APQ\text{ }and\vartriangle \text{ }ABC\] \[\angle APQ\text{ }=\angle ABC\] [As PQ || BC, corresponding angles are equal.] \[\angle PAQ\text{ }=\angle BAC\] [Common angle]...
Solution: As, \[\vartriangle CPQ\text{ }\sim\text{ }\vartriangle CAB\text{ }by\text{ }AA\] criterion for similarity We have, \[CP/AC\text{ }=\text{ }CQ/CB\] \[CP/AC\text{ }=\text{ }4.8/8.4\text{...
Solution: (i) In \[\vartriangle CPQ\text{ }and\text{ }\vartriangle CAB\] \[\angle PCQ\text{ }=\angle APQ\] [As PQ || AB, corresponding angles are equal.] \[\angle C\text{ }=\angle C\] [Common angle]...
Solution: Because, \[\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\text{ }by\text{ }AA\] criterion for similarity So, we have \[AD/AB\text{ }=\text{ }DE/BC\] \[3/8\text{ }=\text{ }DE/4.8\]...
Solution: (i) In \[\vartriangle ABC,\text{ }as\text{ }DE\text{ }||\text{ }BC\] Using BPT, \[AD/DB\text{ }=\text{ }AE/\text{ }EC\] So, \[AD/AB\text{ }=\text{ }AE/AC\] Now, \[AD/AB\text{ }=\text{...
Solution: (i) According to the given question, \[AD/DB\text{ }=\text{ }3/5\] And \[DE\text{ }||\text{ }BC\] Using Basic Proportionality theorem, \[AD/DB\text{ }=\text{ }AE/EC\] \[AE/EC\text{...
i) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords. ii) The locus of the points within a circle which are equidistant...
i) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the given two circles. ii) The locus of the centre of all circles...
i) The locus of points is the space inside and the circumference of the circle with a radius of \[2.5\text{ }cm\]and the centre as the given fixed point. ii) The locus is the space outside and...
i) The locus of the points will be the space inside of the circle whose radius is \[3\text{ }cm\]and centre as the given point. ii) The locus of the points will be the space outside of the circle...
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is \[332\text{ }m\]and centre is the point where the gun is fired.
(i) The locus of the point in a rhombus \[ABCD\]which is equidistant from \[AB\text{ }and\text{ }BC\] will be the diagonal \[BD\]of the rhombus. (ii) The locus of the point in a rhombus...
The locus of the point \[P\]is the circumference of a circle with \[AB\] as diameter and satisfies the condition \[A{{B}^{2}}~=\text{ }A{{P}^{2~}}+\text{ }B{{P}^{2}}\]
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to \[4\text{ }cm.\]
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.
The locus of the centres of all the circles passing through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be a diameter which is the perpendicular bisector of the line joining the...
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and the radius equal to distance between the door handle and the axis of rotation...
The locus of the runner, running around a circular track and always keeping a distance of \[1.5\text{ }m\] from the inner edge will be the circumference of a circle where the radius is equal to the...
Solution: As per the given question, Locus of stone is dropped from the top of tower will be vertical line through the point from which the stone is dropped.
The locus of the moving end of the minute hand of the clock will be a circle whose radius will be the length of the minute hand.
The locus of the centre of wheel, is going straight along the level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.
The locus of points which are at a distance of \[2\text{ }cm\]from a fixed line \[AB\]are a pair of straight lines \[l\text{ }and\text{ }m\]which are parallel to the given line at a distance of...
The locus of a point is at a distance of \[3\text{ }cm\]from a fixed point is circumference of a circle whose radius is \[3\text{ }cm\]and the fixed point is the centre of the circle.
Solution: According to the given question, \[AC\] is the side of a regular octagon, \[\angle AOC\text{ }=\text{ }{{360}^{o}}/\text{ }8\text{ }=\text{ }{{45}^{o}}\] Hence, \[arc\text{ }AC\]subtends...
Solution: (i) \[Arc\text{ }AB\] subtends\[\angle AOB\]at the centre and \[\angle ACB\]at the remaining part of the circle. \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB\]...
Join \[OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS\] Given, \[PQ\text{ }=\text{ }QR\text{ }=\text{ }RS\] So, \[\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS\] [Equal chords subtends equal...
Solution: Join \[OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS\] Given, \[PQ\text{ }=\text{ }QR\text{ }=\text{ }RS\] So, \[\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS\] [Equal chords subtends...
Let ABCD is a cyclic quadrilateral in which\[AB\text{ }||\text{ }DC\] \[AC\text{ }and\text{ }BD\]are its diagonals. Required to prove: \[\left( i \right)\text{ }AD\text{ }=\text{ }BC\] \[\left( ii...
Solution: Join \[AE,\text{ }OB\text{ }and\text{ }OC\] (i) As \[AOD\]is the diameter \[\angle AED\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] But, given \[\angle DEF\text{...
Solution: Let \[ABCD\]be the cyclic trapezium in which \[AB\text{ }||\text{ }DC,\text{ }AC\text{ }and\text{ }BD\]are the diagonals. Required to prove: \[\left( i \right)\text{ }AD\text{ }=\text{...
SOLUTION: Since, \[AB\text{ }=\text{ }DE\text{ }=\text{ }10\text{ }m\] So, in ∆ABC \[\begin{array}{*{35}{l}} DE/AE\text{ }=\text{ }tan\text{ }y\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4} \\...
SOLUTION: Let AB to be the tower of height h meters and let C and D be two points on the level ground such that BC = b meters, BD = a meters, ∠ACB = α, ∠ADB = β. Given, \[\alpha \text{ }+\text{...
Let AD to be the height of the man, AD = 2 m. \[=>\text{ }CE\text{ }=\text{ }\left( 10\text{ }-\text{ }2 \right)\text{ }=\text{ }8\text{ }m\] (i) In ∆CED, \[\begin{array}{*{35}{l}} CE/DE\text{...
SOLUTION: Let AB be the tower of height x metre, surmounted by a vertical flagstaff AD. Let C be a point on the plane such that ∠ACB = α, ∠ACB = β and AD = h. In ∆ABC, \[\begin{array}{*{35}{l}}...
SOLUTION: Let AB to be the vertical tower and C and D be the two points such that CD = 192 m. And let ∠ACB = θ and ∠ADB = α \[\begin{array}{*{35}{l}} tan\text{ }\theta \text{ }=\text{ }5/12 \\...
Since, CA = CB = 15 cm and ∠ACB = 131o Constructing a perpendicular CP from centre C to the chord AB,we get CP bisects ∠ACB as well as chord AB. =>∠ACP = 65.5o In ∆ACP, \[\begin{array}{*{35}{l}}...
SOLUTION: In ∆AMOB, \[\begin{array}{*{35}{l}} cos\text{ }{{30}^{o}}~=\text{ }AO/MO \\ \surd 3/2\text{ }=\text{ }AO/6 \\ AO\text{ }=\text{ }5.20\text{ }m \\ \end{array}\] In ∆BNO,...
SOLUTION: In ∆ADC, \[\begin{array}{*{35}{l}} CD/AD\text{ }=\text{ }tan\text{ }{{42}^{o}} \\ CD\text{ }=\text{ }20\text{ x }0.9004\text{ }=\text{ }18.008\text{ }m \\ \end{array}\] In ∆ADB,...
SOLUTION: In ∆PSB, \[\begin{array}{*{35}{l}} PS/PB\text{ }=\text{ }sin\text{ }{{60}^{o}} \\ PB\text{ }=\text{ }2/\text{ }\surd 3\text{ }=\text{ }1.155\text{ }m \\ \end{array}\] In ∆APQ,...
(i) FIG-1 SOLUTION: In ∆AEB, \[\begin{array}{*{35}{l}} AE/BE\text{ }=\text{ }tan\text{ }{{32}^{o}} \\ AE\text{ }=\text{ }20\text{ x }0.6249\text{ }=\text{ }12.50\text{ }m \\ AD\text{ }=\text{...
Let AB to be the tower of height h meters and let the two points be C and D be such that CD = 30 m, ∠ADE = 45o and ∠ACB = 60o (i) In ∆ADE, \[\begin{array}{*{35}{l}} AE/DE\text{ }=\text{ }tan\text{...
SOLUTION: In ∆AED, \[\begin{array}{*{35}{l}} AE/\text{ }DE\text{ }=\text{ }tan\text{ }{{22}^{o}} \\ AE\text{ }=\text{ }DE\text{ }tan\text{ }{{22}^{o}}~=\text{ }15\text{ }x\text{ }0.404\text{...
Let AB and CD be the two towers of height h m each and let P be a point in the roadway BD such that BD = 150 m, ∠APB = 60o and ∠CPD = 30o In ∆ABP, \[\begin{array}{*{35}{l}} AB/BP\text{ }=\text{...
Let AB to be the lighthouse and the two ships be C and D such that ∠ADB = 36o and ∠ACB = 48o In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{ }{{48}^{o}} \\ BC\text{ }=\text{...
Let AB to be the building of height h meters. and let the two points be C and D be such that CD = 40 m, ∠ADB = 30o and ∠ACB = 45o In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{...
Let AB to be the height of the tree, h m. and let the two points be C and D be such that CD = 20 m, ∠ADB = 30o and ∠ACB = 60o In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{...
In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{ }{{45}^{o}}~=\text{ }1 \\ =>\text{ }BC\text{ }=\text{ }AB\text{ }=\text{ }X \\ \end{array}\] In ∆ABD,...
Let the height of the tower to be ‘h’ m. (i) Since, \[\begin{array}{*{35}{l}} \theta \text{ }=\text{ }{{45}^{o}} \\ tan\text{ }{{45}^{o}}~=\text{ }\left( h\text{ }-\text{ }1.6 \right)/\text{ }20 ...
Let the length of the rope to be x meters. \[\begin{array}{*{35}{l}} sin\text{ }{{30}^{o}}~=\text{ }60/x \\ {\scriptscriptstyle 1\!/\!{ }_2}\text{ }=\text{ }60/x \\ x\text{ }=\text{ }120\text{ }m ...
Let one of the persons be A , at a distance of ‘x’ meters and the second person be B at a distance of ‘y’ meters from the foot of the tower. The angle of elevation of A is 30o...
Let the height upto which the ladder reaches as ‘h’ meters. the angle of elevation is 68o \[\begin{array}{*{35}{l}} =>\text{ }tan\text{ }{{68}^{o}}~=\text{ }h/\text{ }2.4 \\ 2.475\text{ }=\text{...
Let the height of the tower to be h meters. the angle of elevation is 60o => \[\begin{array}{*{35}{l}} tan\text{ }{{60}^{o}}~=\text{ }h/160 \\ \surd 3\text{ }=\text{ }h/160 \\ h\text{ }=\text{...
Let the length of the shadow of the tree to be x meters. Therefore, the height of the tree = √3 x meters If θ is the angle of elevation of the sun, \[\begin{array}{*{35}{l}} =>\text{ }tan\text{...
According to ques, \[~1/x\text{ }-\text{ }1/\left( x-2 \right)\text{ }=\text{ }3\] \[\left( x\text{ }-\text{ }2\text{ }\text{ }x \right)/\text{ }\left( x\left( x\text{ }-\text{ }2 \right)...
According to ques, \[m\text{ }+\text{ }5\text{ }=\text{ }0\text{ }and\text{ }n\text{ }\text{ }1\text{ }=\text{ }0\] hence, \[m\text{ }=\text{ }-5\text{ }and\text{ }n\text{ }=\text{ }1\] putting...
According to the given question, \[p\text{ }\text{ }15\text{ }=\text{ }0\] And \[2{{x}^{2}}~+\text{ }px\text{ }+\text{ }25\text{ }=\text{ }0\] Thus, \[p\text{ }=\text{ }15\] Using\[p\]in the...
According to the given equation, \[{{x}^{2}}~+\text{ }\left( 3\text{ }\text{ }2a \right)x\text{ }\text{ }6a\text{ }=\text{ }0\] By putting \[x\text{ }=\text{ }-3\]we get \[{{\left( -3...
According to the given question, \[8{{x}^{2~}}+\text{ }mx\text{ }+\text{ }15\text{ }=\text{ }0\] One of the roots is\[~{\scriptscriptstyle 3\!/\!{ }_4},\]and it satisfies the given equation So,...
According to the given question, \[2x\text{ }\text{ }3\text{ }=\text{ }\surd (2{{x}^{2}}~\text{ }2x\text{ }+\text{ }21)\] Squaring on both sides, we get \[{{\left( 2x\text{ }\text{ }3...
![\[\left( \mathbf{i} \right)\text{ }\mathbf{3}\surd \mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }\text{ }\surd \mathbf{2}\text{ }=\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-22ead8085663c05474ddb007572bafdc_l3.png "Rendered by QuickLaTeX.com")
According to the given question, \[3\surd 2{{x}^{2}}~\text{ }5x\text{ }\text{ }\surd 2\text{ }=\text{ }0\] Or, \[3\surd 2{{x}^{2}}~\text{ }6x\text{ }+\text{ }x\text{ }\text{ }\surd 2\text{ }=\text{...
![\[\left( \mathbf{ii} \right)\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{2}\surd \mathbf{6x}\text{ }+\text{ }\mathbf{2}\text{ }=\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d5146bd91c51cf5cda3cc06dbd2ed9d1_l3.png "Rendered by QuickLaTeX.com")
According to the given equation, \[3{{x}^{2}}~\text{ }2\surd 6x\text{ }+\text{ }2\text{ }=\text{ }0\] Or, \[3{{x}^{2}}~\text{ }\surd 6x\text{ }\text{ }\surd 6x\text{ }+\text{ }2\text{ }=\text{ }0\]...
![\[\left( \mathbf{i} \right)\text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{5} \right)\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{5} \right)\text{ }=\text{ }\mathbf{24}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-85409799f13ae2131936b8199e40aef0_l3.png "Rendered by QuickLaTeX.com")
According to the given question, \[\left( x\text{ }+\text{ }5 \right)\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }24\] Or, \[{{x}^{2}}~\text{ }25\text{ }=\text{ }24\] Or,...
Let \[\left( x\text{ }\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring on both sides \[{{\left( x\text{ }\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\] or,...
let \[\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring on both sides \[{{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\]...
Let, \[~\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring both sides \[{{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\]...
According to ques, \[{{x}^{4}}~\text{ }2{{x}^{2}}~\text{ }3\text{ }=\text{ }0\] Let’s take \[{{x}^{2}}~=\text{ }y\] Then, the equation becomes \[{{y}^{2}}~\text{ }2y\text{ }\text{ }3\text{ }=\text{...
Solution: So, on comparison we get \[x\text{ }=\text{ }36.\]
Solution: As per the given question,
According to ques, \[~2{{x}^{4}}~\text{ }5{{x}^{2}}~+\text{ }3\text{ }=\text{ }0\] Let \[{{x}^{2}}~=\text{ }y\] The, \[2{{y}^{2}}~\text{ }5y\text{ }+\text{ }3\text{ }=\text{ }0\] \[2{{y}^{2}}~\text{...
Solution: Now, On comparison, we get \[-28\text{ }-\text{ }3x\text{ }=\text{ }10\] \[3x\text{ }=\text{ }-38\] \[x\text{ }=\text{ }-38/3\] And, \[20\text{ }-\text{ }3y\text{ }=\text{ }-8\] \[3y\text{...
Solution: According to the given question, On comparison, we get \[2x\text{ }+\text{ }3x\text{ }=\text{ }5\] And \[2y\text{ }+\text{ }4y\text{ }=\text{ }12\] \[5x\text{ }=\text{ }5\text{ }and\text{...
According to ques, \[~x\text{ }+\text{ }4/x\text{ }=\text{ }-4\] or, \[\left( {{x}^{2~}}+\text{ }4 \right)/\text{ }x\text{ }=\text{ }-4\] Or, \[{{x}^{2}}~+\text{ }4\text{ }=\text{ }-4x\] or,...
Solution: \[\left( i \right)\text{ }A\text{ }\left( BA \right)\] \[\left( ii \right)\text{ }\left( AB \right)\text{ }B\]
Solution: According to the given ques, On comparison, we get \[a\text{ }+\text{ }1\text{ }=\text{ }5\Rightarrow a\text{ }=\text{ }4\] \[b\text{ }+\text{ }2\text{ }=\text{ }0\Rightarrow b\text{...
According to ques, \[~{{x}^{2}}~\text{ }11/4\text{ }x\text{ }+\text{ }15/8\text{ }=\text{ }0\] Taking L.C.M , \[\left( 8{{x}^{2}}~\text{ }22x\text{ }+\text{ }15 \right)/\text{ }8\text{ }=\text{ }0\]...
Solution: According to the given question, \[3A\text{ }x\text{ }M\text{ }=\text{ }2B\] Suppose the order of the \[matrix\text{ }M\text{ }be\text{ }\left( a\text{ }x\text{ }b \right)\] Now, we know...
According to ques, \[{{a}^{2}}{{x}^{2}}~\text{ }{{b}^{2}}~=\text{ }0\] or, \[{{\left( ax \right)}^{2}}~\text{ }{{b}^{2}}~=\text{ }0\] Or, \[\left( ax\text{ }+\text{ }b \right)\left( ax\text{ }\text{...
Solution: As per the given question,
Solution: (i) Suppose, the order of the matrix be \[a\text{ }x\text{ }b\] We know that Hence, for product of matrices to be possible \[a\text{ }=\text{ }2\] And, form the order of the resultant...
According to ques, \[{{\left( 2x\text{ }+\text{ }3 \right)}^{2}}~=\text{ }81\] Taking square root both sides, \[2x\text{ }+\text{ }3\text{ }=\text{ }\pm \text{ }9\] Or, \[2x\text{ }=\text{ }\pm...
Solution: According to the given question, \[{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }25\] And, \[-2{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }-2\] \[\left( i \right)\text{ }x,\text{ }y~\in ~W\text{ }\left(...
Solution: On comparison, we get \[3x\text{ }+\text{ }18\text{ }=\text{ }15\] And \[12x\text{ }+\text{ }77\text{ }=\text{ }10y\] \[3x\text{ }=\text{ }-3\] And \[y\text{ }=\text{ }\left( 12x\text{...
3,x -3/2According to ques, \[4{{x}^{2}}~+\text{ }6x\text{ }+\text{ }x\text{ }\text{ }3\text{ }+\text{ }3x\text{ }+\text{ }9\text{ }=\text{ }0\] Or, \[4{{x}^{2}}~+\text{ }10x\text{ }+\text{ }6\text{ }=\text{...
Solution: On comparison, we get \[6x\text{ }-\text{ }10\text{ }=\text{ }8\] And \[-2x\text{ }+\text{ }14\text{ }=\text{ }4y\] \[6x\text{ }=\text{ }18\] And \[y\text{ }=\text{ }\left( 14\text{...
Solution: \[{{A}^{2}}~+\text{ }BC\]
Solution: \[\left( i \right)\text{ }A\left( B\text{ }+\text{ }C \right)\] \[AB\text{ }+\text{ }AC\] So, \[A\left( B\text{ }+\text{ }C \right)\text{ }=\text{ }AB\text{ }+\text{ }AC\] \[\left( ii...
Solution: \[{{A}^{2}}\] Given, \[~{{A}^{2~}}=\text{ }I\] On comparison, we get \[1\text{ }+\text{ }a\text{ }=\text{ }1\] \[a\text{ }=\text{ }0\] And, \[-1\text{ }+\text{ }b\text{ }=\text{ }0\]...
let the present ages of father and his son be \[~x\text{ }years\text{ }and\text{ }\left( 45\text{ }\text{ }x \right)\text{ }years\] hence five years ago, Father’s age \[=\text{ }\left( x\text{...
Solution: \[{{B}^{2}}\] \[{{B}^{2}}~=\text{ }B\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}A\] \[{\scriptscriptstyle 1\!/\!{ }_2}A\text{ }=\text{ }{{B}^{2}}-\text{ }B\] \[A\text{ }=\text{...
Solution: According to the given ques, \[\left( i \right)\text{ }\left( A\text{ }+\text{ }B \right)\] \[So,\text{ }{{\left( A\text{ }+\text{ }B \right)}^{2}}~=\text{ }\left( A\text{ }+\text{ }B...
According to ques, \[S\text{ }=\text{ }n\left( n\text{ }+\text{ }1 \right)\] Also, \[S\text{ }=\text{ }420\] So, \[n\left( n\text{ }+\text{ }1 \right)\text{ }=\text{ }420\] Or, \[{{n}^{2}}~+\text{...
Solution: \[\left( \mathbf{i} \right)\text{ }\mathbf{AB}\text{ }\] \[\left( ii \right)\text{ }{{\mathbf{A}}^{\mathbf{2}}}-\text{ }\mathbf{AB}\text{ }+\text{ }\mathbf{2B}\]
Let the speed of the second train be \[~x\text{ }km/hr.\] Then, the speed of the first train is \[\left( x\text{ }+\text{ }5 \right)\text{ }km/hr\] Let O be the position of the railway station,...
Solution: \[\left( \mathbf{i} \right)\text{ }\mathbf{A}\text{ }-\text{ }\mathbf{B}\text{ }\] \[\text{ }\left( \mathbf{ii} \right)\text{ }{{\mathbf{A}}^{\mathbf{2}~}}\]
Solution: $BA$ \[{{M}^{2}}\] So, \[BA\text{ }={{M}^{2}}\] On comparison, we get \[-2b\text{ }=\text{ }-2\] \[b\text{ }=\text{ }1\] And, \[a\text{ }=\text{ }2\]
Solution: \[{{M}^{2}}\] \[2M\text{ }+\text{ }3I\] Hence, \[{{M}^{2}}~=\text{ }2M\text{ }+\text{ }3I\]
Let the usual speed of the plane to be \[x\text{ }km/hr\] The distance to travel \[=\text{ }1500km\] since, Time = Distance/ Speed As the ques suggests, \[{{x}^{2}}~+\text{ }250x\text{ }\text{...
Solution: \[{{A}^{2}}\] $AC$ $5B$ \[{{A}^{2}}~+\text{ }AC\text{ }-\text{ }5B\text{ }=\]
Solution: \[{{A}^{2}}~=\text{ }A\text{ }x\text{ }A,\text{ }\]isn’t possible because the number of columns isn’t equal to its number of rows in matrix A.
let the original number of persons to be x. According to ques, Total money which was divided is \[=\text{ }Rs\text{ }6500\] Each person’s share is \[=\text{ }Rs\text{ }6500/x\] Then, as the question...
Solution: \[\left( i \right)\text{ }AB\] \[\left( ii \right)\text{ }BA\]
Solution: According to the given ques, \[\left( i \right)\text{ }\left( AB \right)\] \[\left( AB \right)\text{ }C\] \[\left( ii \right)\text{ }BC\] \[A\text{ }\left( BC \right)\] So, \[\text{...
According to ques, Distance \[=\text{ }400\text{ }km\] Average speed of the airplane \[=\text{ }x\text{ }km/hr\] Also, speed while returning \[=\text{ }\left( x\text{ }+\text{ }40 \right)\text{...
Solution: (i) On comparison, we get \[5x\text{ }-\text{ }2\text{ }=\text{ }8\] \[5x\text{ }=\text{ }10\] \[x\text{ }=\text{ }2\] And, \[20\text{ }+\text{ }3x\text{ }=\text{ }y\] \[20\text{ }+\text{...
Let the number of people staying overnight as x. According to ques, total hotel bill \[~=\text{ }Rs\text{ }4800\] Now,hotel bill for each person \[=\text{ }Rs\text{ }4800/x\] therefore,...
Solution: \[{{A}^{2}}~\] \[{{A}^{2}}~=\text{ }B\] On comparison, we get \[4x\text{ }=\text{ }16\] \[x\text{ }=\text{ }4\] And, \[1\text{ }=\text{ }-y\] \[y\text{ }=\text{ }-1\]
Solution: (i) \[{{A}^{2}}\] (ii) \[~{{B}^{2}}\] \[{{B}^{2}}A\]
According to ques, Number of articles \[=\text{ }x\] And, the total cost of articles \[=\text{ }Rs\text{ }600\] Again, (i) Cost of one article \[=\text{ }Rs\text{ }600/x\] (ii) also,...
Solution: (i) AI = (ii) IB=
Solution: According to the given question (i) (ii)
(iii) According to the question, \[4x\text{ }+\text{ }1728\text{ }=\text{ }{{x}^{2}}~+\text{ }16x\] Or, \[{{x}^{2}}~+\text{ }12x\text{ }\text{ }1728\text{ }=\text{ }0\] Or, \[{{x}^{2}}~+\text{...
Solution: The product of the given matrices isn’t possible as per the rule the number of columns in the first matrix isn’t equal to the number of rows in the second matrix.
Solution: \[=\text{ }\left[ 6\text{ }+\text{ }0 \right]\text{ }=\text{ }\left[ 6 \right]\] \[=\text{ }\left[ -2+2\text{ }3-8 \right]\text{ }=\text{ }\left[ 0\text{ }-5 \right]\]
According to ques, Speed of car = \[x\text{ }km/hr\] Speed of train = \[\left( x\text{ }+\text{ }16 \right)\text{ }km/hr\] Time = \[Distance/\text{ }Speed\] (i)Time taken by the car to reach town B...
(ii) Solution: (i) \[2A\text{ }+\text{ }B\] (ii)
Solution: According to the given question, the matrix is
Solution: \[\left( i \right)\text{ }2A\text{ }-\text{ }3B\text{ }+\text{ }C\] \[\left( ii \right)\text{ }A\text{ }+\text{ }2C\text{ }-\text{ }B\]
(ii) Solution: From L.H.S, we have \[3\left[ 4\text{ }x \right]\text{ }+\text{ }2\left[ y\text{ }-3 \right]\] \[=\text{ }\left[ 12\text{ }3x \right]\text{ }+\text{ }\left[ 2y\text{ }-6 \right]~\]...
(I) (ii) Solution: According to the given ques, (i) (ii)
(ii) Solution: (i) \[3\left[ 5\text{ }-2 \right]\text{ }=\text{ }\left[ 3\times 5\text{ }3x-2 \right]\text{ }=\text{ }\left[ 15\text{ }-6 \right]\] (ii)
According to the given ques, Hence, the common ratio is \[3/5.\]
. According to the given question, G.P: \[2\text{ }+\text{ }6\text{ }+\text{ }18\text{ }+\text{ }\ldots ..\] Here, \[a\text{ }=\text{ }2\] And \[r\text{ }=\text{ }6/2\text{ }=\text{ }3\] Also given,...
According to the given question, G.P: \[3,\text{ }6,\text{ }12,\text{ }\ldots .,\text{ }1536\] So, \[a\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }1536\] and \[r\text{ }=\text{ }6/3\text{ }=\text{...
According to the given question, For a G.P., \[r\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }486\] and \[{{S}_{n}}~=\text{ }728\] \[1458\text{ }-\text{ }a\text{ }=\text{ }728\text{ }x\text{ }2\text{...
According to the given question, \[{{t}_{4}}~=\text{ }1/27\] and \[{{t}_{7~}}=\text{ }1/729\] We know that, \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4~}}=\text{...
Amount spent on \[{{1}^{st}}~day\text{ }=\text{ }Rs\text{ }10\] Amount spent on \[{{2}^{nd}}~day\text{ }=\text{ }Rs\text{ }20\] Amount spent on \[{{3}^{rd}}~day\text{ }=\text{ }Rs\text{ }40\]...
\[First\text{ }term\text{ }\left( a \right)\text{ }of\text{ }a\text{ }G.P\text{ }=\text{ }27\] And, \[{{8}^{th}}~term\text{ }=\text{ }{{t}_{8}}~=\text{ }a{{r}^{8\text{ }-\text{ }1}}~=\text{ }1/81\]...
According to the given question, G.P: \[1\text{ }+\text{ }4\text{ }+\text{ }16\text{ }+\text{ }64\text{ }+\text{ }\ldots \ldots ..\] Here, \[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{...
(i) (ii) Solution: (i) According to the given question Here, \[a\text{ }=\text{ }\left( x\text{ }+\text{ }y \right)/\text{ }\left( x\text{ }-\text{ }y \right)\] And \[~r\text{ }=\text{ }1/\left[...
(i) According to the given question G.P: \[1\text{ }-\text{ }1/2\text{ }+\text{ }1/4\text{ }-\text{ }1/8\text{ }+\text{ }\ldots \ldots ..\text{ }to\text{ }9\text{ }terms\] Here, \[a\text{ }=\text{...
(i) According to the given question G.P: \[1\text{ }+\text{ }3\text{ }+\text{ }9\text{ }+\text{ }27\text{ }+\text{ }\ldots \ldots \ldots .\text{ }to\text{ }12\text{ }terms\] Here, \[a\text{ }=\text{...
The first term of the G.P. be A and its common ratio be R. Hence, \[{{p}^{th}}~term\text{ }=\text{ }a\text{ }\Rightarrow \text{ }A{{R}^{p\text{ }-\text{ }1}}~=\text{ }a\] \[{{q}^{th}}~term\text{...
According to the given question, G.P. \[1/27,\text{ }1/9,\text{ }1/3,\text{ }\ldots \ldots ,\text{ }81\] Here, \[a\text{ }=\text{ }1/27,\] \[common\text{ }ratio\text{ }\left( r \right)\text{...
Given series: \[2/27,\text{ }2/9,\text{ }2/3,\text{ }\ldots \ldots .,\text{ }162\] Here, \[a\text{ }=\text{ }2/27\] \[r\text{ }=\text{ }\left( 2/9 \right)\text{ }/\text{ }\left( 2/27 \right)\]...
(iv) 
(iii) $\sin \left(28^{\circ}+\mathrm{A}\right)=\sin \left[90^{\circ}-\left(62^{\circ}-\mathrm{A}\right)\right]=\cos \left(62^{\circ}-\mathrm{A}\right)$ (iv)
Given series: \[\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32\] Here, \[a\text{ }=\text{ }\surd 2\] \[r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2\]...
(ii) 
(i) $\tan \left(55^{\circ}+x\right)=\tan \left[90^{\circ}-\left(35^{\circ}-x\right)\right]=\cot \left(35^{\circ}-x\right)$ (ii) $\sec \left(70^{\circ}-\theta\right)=\sec...
(viii) 
(vii) $3 \cos 80^{\circ} \operatorname{cosec} 10^{\circ}+2 \cos 59^{\circ} \operatorname{cosec} 31^{\circ}$ $=3 \cos (90-10)^{0} \operatorname{cosec} 10^{\circ}+2 \cos (90-31)^{0}...
According to the given question, Product of \[{{3}^{rd}}~and\text{ }{{8}^{th}}\] terms of a G.P. is \[243\] The general term of a G.P. First term \[a\] And Common ratio \[r\]is given by,...
(vi) 
(v) $\sin 27^{\circ} \sin 63^{\circ}-\cos 63^{\circ} \cos 27^{\circ}$ $=\sin (90-63)^{0} \sin 63^{\circ}-\cos 63^{\circ} \cos (90-63)^{\circ}$ $=\cos 63^{\circ} \sin 63^{\circ}-\cos 63^{\circ} \sin...
(iv) 
(iii) (iv) $\cos 40^{\circ} \operatorname{cosec} 50^{\circ}+\sin 50^{\circ}$ sec $40^{\circ}$ $=\cos (90-50)^{0} \operatorname{cosec} 50^{\circ}+\sin (90-50)^{0} \sec 40^{\circ}$ $=\sin 50^{\circ}...
(ii) 
(i) (ii) 1+1=2
According to the given question, \[{{t}_{1}}~=\text{ }2\text{ }and\text{ }{{t}_{3}}~=\text{ }8\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{1}}~=\text{...
Given, \[{{t}_{4}}~=\text{ }1/18\text{ }and\text{ }{{t}_{7}}~=\text{ }-1/486\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4}}~=\text{ }a{{r}^{4\text{ }-\text{...
According to the given question, \[{{t}_{5}}~=\text{ }81\text{ }and\text{ }{{t}_{2}}~=\text{ }24\] We know that, General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So,...
, show that: 
(ii) If 
, show that: 
(i) Since, $2 \sin \mathrm{A}-1=0$ Therefore, $\sin A=1 / 2$ since, $\sin 30^{\circ}=1 / 2$ Hence, $\mathrm{A}=30^{\circ}$ LHS = $\sin 3 A=\sin 3\left(30^{\circ}\right)=\sin 30^{\circ}=1$...
$\tan A=n \tan B$ $n=\tan A / \tan B$ And, $\sin A=m \sin B$ $\mathrm{m}=\sin \mathrm{A} / \sin \mathrm{B}$ Substituting RHS in $\mathrm{m}$ and $\mathrm{n}$ $m^{2}-1 / n^{2}-1$...
, show that: 
RHS, $\left(p^{2}-1\right) /\left(p^{2}+1\right)$ $=\frac{(\sec A+\tan A)^{2}-1}{(\sec A+\tan A)^{2}+1}$ $=\frac{\sec ^{2} A+\tan ^{2} A+2 \tan A \sec A-1}{\sec ^{2} A+\tan ^{2} A+2 \tan A \sec...
and 
, show that: 
LHS = $a^{2} / x^{2}-b^{2} / y^{2}$ $=\frac{a^{2}}{a^{2} \cos ^{2} \theta}-\frac{b^{2}}{b^{2} \cot ^{2} \theta}$ $=\frac{1}{\cos ^{2} \theta}-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$...
LHS = $q\left(p^{2}-1\right)=(\sec A+\operatorname{cosec} A)\left[(\sin A+\cos A)^{2}-1\right]$ $=(\sec A+\operatorname{cosec} A)\left[\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A-1\right]$ $=(\sec...
Solution: In the given G.P. First term, \[a\text{ }=\text{ }-10\] Common ratio, \[r\text{ }=\text{ }\left( 5/\surd 3 \right)/\text{ }\left( -10 \right)\text{ }=\text{ }1/\left( -2\surd 3 \right)\]...
(xvi) 
(xv) LHS = $\sec ^{4} \mathrm{~A}\left(1-\sin ^{4} \mathrm{~A}\right)-2 \tan ^{2} \mathrm{~A}$ $=\sec ^{4} \mathrm{~A}\left(1-\sin ^{2} \mathrm{~A}\right)\left(1+\sin ^{2} \mathrm{~A}\right)-2 \tan...
$(1+\tan A+\sec A)(1+\cot A-\operatorname{cosec} A)$ $=1+\cot A-\operatorname{cosec} A+\tan A+1-\sec A+\sec A+\operatorname{cosec} A-\operatorname{cosec} A \sec A$ $=2+\cos \mathrm{A} / \sin...
(xiv) 
LHS = = RHS (xiv) LHS = = RHS
According to the given question, The mid-point of \[\left( 2a,\text{ }4 \right)\text{ }and\text{ }\left( -2,\text{ }2b \right)\text{ }is\text{ }\left( 1,\text{ }2a\text{ }+\text{ }1 \right)\] By...
(xii) 
LHS = = RHS (xii) LHS = = RHS
(x) 
LHS= = RHS (x) LHS =
(viii) 
(vii) LHS =$=(\sin A /(1-\cos A))-\cot A$ Since, $\cot A=\cos A / \sin A$ $=\left(\sin ^{2} A-\cos A+\cos ^{2} A\right) /(1-\cos A) \sin A$ $=(1-\cos A) /(1-\cos A) \sin A$ $=1 / \sin \mathrm{A}$...
(vi) 
(v) LHS= cot A/ (1 – tan A) + tan A/ (1 – cot A) = RHS (vi) LHS= cos A/ (1 + sin A) + tan A = RHS
According to the given question, The mid-point of the line-segment joining \[\left( 4a,\text{ }2b\text{ }-\text{ }3 \right)\text{ }and\text{ }\left( -4,\text{ }3b \right)\text{ }is\text{ }\left(...
(iv) 
(iii) LHS= 1 – sin2 A/ (1 + cos A) = RHS (iv) LHS= (1 – cos A)/ sin A + sin A/ (1 – cos A) = RHS
By the centroid of a triangle formula, we get the co- ordinates of the centroid of triangle \[ABC\]as \[=\text{ }\left( 5/3,\text{ }1 \right)\]
(i) Taking LHS, $1 /(\cos A+\sin A)+1 /(\cos A-\sin A)$ (ii) Taking LHS, $\operatorname{cosec} A-\cot A$ $=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$ $=\frac{1-\cos A}{\sin A}$ $=\frac{1-\cos A}{\sin...
(i) \[Radius\text{ }AC\text{ }=\text{ }\surd [\text{ }{{\left( 3\text{ }+\text{ }2 \right)}^{2}}~+\text{ }{{\left( -7\text{ }-\text{ }5 \right)}^{2}}~]\] \[=\text{ }\surd [\text{ }({{5}^{2}}~+\text{...
(iii) $\tan 17^{\circ} 27^{\prime}=\tan \left(17^{\circ} 24^{\prime}+3^{\prime}\right)=0.3134+0.0010=0.3144$