\[\mathbf{sin}\text{ }\mathbf{P},~\mathbf{sin}\text{ }\mathbf{R},\text{ }\mathbf{sec}\text{ }\mathbf{P}\text{ }\mathbf{and}\text{ }\mathbf{sec}\text{ }\mathbf{R}.\]

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

\[sec\text{ }R\text{ }=\text{ }PR/\text{ }RQ\]

\[\Rightarrow sec\text{ }R\text{ }=\text{ }5/3\]