Solution: Given that $Z=\left(i^{25}\right)^{3}$ $\begin{array}{l} =\dot{i}^{75} \\ =\mathrm{i}^{74} \cdot \mathrm{i} \\ =\left(\mathrm{i}^{2}\right)^{37} \cdot \mathrm{i} \\ =(-1)^{37} \cdot...
Solution: It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$ In which, $\begin{array}{l} \left.|Z|=\text { modulus of complex number...
Solution: It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$ In which, $\begin{array}{l} \left.|Z|=\text { modulus of complex number...
Solution: It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$ In which, $\begin{array}{l} \left.|Z|=\text { modulus of complex number...
Solution: Given that $\begin{array}{l} z_{1}=\bar{z}_{2} \\ z_{3}=\bar{z}_{4} \end{array}$ It is known that $\arg \left(\mathrm{z}_{1} / \mathrm{z}_{2}\right)=\arg \left(\mathrm{z}_{1}\right)-\arg...
Solution: Given that $Z=\sin \pi / 5+i(1-\cos \pi / 5)$ Using the formula, $\begin{array}{l} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta \end{array}$ Therefore,...
Solution: Let $A B C$ be the metallic cone, $DECB$ is the required frustum Let the two radii of the frustum be$\mathrm{DO}^{\prime}=\mathrm{r}_{2}$ and $\mathrm{BO}=\mathrm{r}_{1}$From $\triangle...
Given, r1 = 20 cm, r2 = 8 cm and h = 16 cm \[\therefore Volume\text{ }of\text{ }the\text{ }frustum\text{ }=\text{ }\left( \right)\times \pi \times h\left( r12+r22+r1r2 \right)\] It is given that...
Given, For the lower roundabout end, span $(r_1)$ = 10 cm For the upper roundabout end, span $(r_2)$ = 4 cm Inclination tallness (l) of frustum = 15 cm Presently, The space of material to be...
Given, Inclination tallness (l) = 4 cm Perimeter of upper roundabout finish of the frustum = 18 cm \[\therefore 2\pi r_1\text{ }=\text{ }18\] Or on the other hand, $r_1$ = 9/π Likewise, periphery of...
Sweep $(r_1)$ of the upper base = 4/2 = 2 cm Sweep $(r_2)$ of lower the base = 2/2 = 1 cm Tallness = 14 cm Presently, Capacity of glass = Volume of frustum of cone Thus, Capacity of glass = \[\left(...