\[\mathbf{sin}\text{ }\mathbf{P},~\mathbf{sin}\text{ }\mathbf{R},\text{ }\mathbf{sec}\text{ }\mathbf{P}\text{ }\mathbf{and}\text{ }\mathbf{sec}\text{ }\mathbf{R}.\]
Solution:
Given:
\[\vartriangle PQR~\]is right-angled at Q.
\[PQ\text{ }=\text{ }4cm\]
\[RQ\text{ }=\text{ }3cm\]
Required to find: \[sin\text{ }P,\text{ }sin\text{ }R,\text{ }sec\text{ }P,\text{ }sec\text{ }R\]
Given \[\vartriangle PQR,\]
By using Pythagoras theorem to \[\vartriangle PQR,\], we get
\[P{{R}^{2}}~=\text{ }P{{Q}^{2}}~+R{{Q}^{2}}\]
Substituting the respective values,
\[P{{R}^{2}}~=\text{ }{{4}^{2}}~+{{3}^{2}}\]
\[P{{R}^{2}}~=\text{ }16\text{ }+\text{ }9\]
\[P{{R}^{2}}~=\text{ }25\]
\[PR\text{ }=~\surd 25\]
\[PR\text{ }=\text{ }5\]
\[\Rightarrow Hypotenuse\text{ }=5\]
By definition,
sin P = Perpendicular side opposite to angle P/ Hypotenuse
\[sin\text{ }P\text{ }=\text{ }RQ/\text{ }PR\]
\[\Rightarrow sin\text{ }P\text{ }=\text{ }3/5\]
And,
sin R = Perpendicular side opposite to angle R/ Hypotenuse
\[sin\text{ }R\text{ }=\text{ }PQ/\text{ }PR\]
\[\Rightarrow sin\text{ }R\text{ }=\text{ }4/5\]
And,
\[sec\text{ }P=1/cos\text{ }P\]
secP = Hypotenuse/ Base side adjacent to ∠P
\[sec\text{ }P\text{ }=\text{ }PR/\text{ }PQ\]
\[\Rightarrow sec\text{ }P\text{ }=\text{ }5/4\]
Now,
\[sec\text{ }R\text{ }=\text{ }1/cos\text{ }R\]
secR = Hypotenuse/ Base side adjacent to ∠R
\[sec\text{ }R\text{ }=\text{ }PR/\text{ }RQ\]
\[\Rightarrow sec\text{ }R\text{ }=\text{ }5/3\]