Question

Given $15\cot A=8$, find $\sin A$ and $\sec A.$

Answer 1

Given,

$15\cot A=8$

$\cot A=\frac{8}{15}$

$=>$ $\tan A=\frac{15}{8}$-------($\tan A=\frac{1}{\cot A}$)

We know that,

$\tan \theta =\frac{oppositeSide}{adjacentSide}$

Consider the attached figure, triangle$ABC$

From Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=8^2+{15}^2=64+225=289$

$AC=17$

$\cos A=\frac{adjacentSide}{Hypotenuse}=\frac{AB}{AC}=\frac{8}{17}$

$\sec A=\frac{1}{\cos A}=\frac{1}{\frac{8}{17}}\frac{17}{8}$

$\sin A=\frac{oppositeSide}{Hypotenuse}=\frac{BC}{AC}=\frac{15}{17}$