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
\(\text{ sin A} = \frac{\text{Side}\ \text{ Opposite}\ \text{ to}\ ∠A }{\text{Hypotenuse}}\)

\(\frac{BC}{AC}= \frac{15}{17}\)

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

\(\frac{AC}{AB}=\frac{17}{8}\)