Question

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

Answer 1

Consider a right-angled triangle ABC, with the right angle at B. Given that \( \text{cot A} = \frac{8}{15} \).

We know that in a right-angled triangle, \( \text{cot A} = \frac{\text{Side Adjacent to ∠A}}{\text{Side Opposite to ∠A}} = \frac{AB}{BC} \). Therefore, \( \frac{AB}{BC} = \frac{8}{15} \).

Let AB = 8k and BC = 15k, where k is a positive integer. Using the Pythagorean theorem in \( Δ \)ABC, we have \( \text{AC}^ 2 = \text{AB}^ 2 + \text{BC}^ 2 \). Substituting the values, \( \text{AC}^ 2 = (8k)^2 + (15k)^2 = 64k^2 + 225k^2 = 289k^2 \). Thus, AC = 17k.

Now, we can find \( \text{sin A} \) and \( \text{sec A} \).

The formula for \( \text{sin A} \) is \( \frac{\text{Side Opposite to ∠A}}{\text{Hypotenuse}} \). Therefore, \( \text{sin A} = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17} \).

The formula for \( \text{sec A} \) is \( \frac{\text{Hypotenuse}}{\text{Side Adjacent to ∠A}} \). Therefore, \( \text{sec A} = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8} \).