Question

Given 15 cot A = 8. Find sin A and sec A.

Answer 1

Consider a right-angled triangle, right-angled at B.

$\text{ cot A} = \frac{\text{Side}\ \text{ Adjacent}\ \text{ to}\ ∠A }{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠A }$
$\frac{AB}{BC}$
It is given that,
cot A =$\frac{8}{15}$

$\frac{AB}{BC}=\frac{8}{15}$

Let AB be 8k. Therefore, BC will be 15k, where k is a positive integer.
Applying Pythagoras theorem in $Δ$ABC, we obtain.
$\text{AC}^ 2 =\text{ AB}^ 2 +\text{ BC}^ 2$
$= (8k)^ 2 + (15k) ^2$
$= 64k ^2 + 225k ^2$
$= 289k^ 2$
AC = 17k
$\text{ sin A} = \frac{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠A }{\text{Hypotenuse}}$

$\frac{BC}{AC}= \frac{15}{17}$

$\text{ sec A} = \frac{\text{Hypotenuse}}{\text{Side}\ \text{ Adjacent}\ \text{ to}\ ∠A }$

$\frac{AC}{AB}=\frac{17}{8}$