Question

If sin A = \(\frac{3}{4}\), calculate cos A and tan A.

Answer 1

Consider a right-angled triangle \(ΔABC\), with the right angle at point B.

It is given that,
sin A = \( \frac{3}{4} \)

\( \frac{BC}{AC} = \frac{3}{4} \)

Let BC = 3k and AC = 4k, where k is a positive integer.
Applying the Pythagorean theorem to \(ΔABC\), we get:
\( \text{AC} ^2 = \text{AB} ^2 + \text{BC} ^2 \)
\( (4k) ^2 = \text{AB}^ 2 + (3k)^ 2 \)
\( 16k ^2 - 9k^ 2 = \text{ AB}^ 2 \)
\( 7k ^2 = \text{AB}^ 2 \)
\( AB = \sqrt{7}k \).

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

\( \frac{AB}{AC} = \frac{ \sqrt{7}k}{4k} = \frac{ \sqrt{7}}{4} \)

\( \text{ tan A} = \frac{\text{Opposite Side to } ∠A }{\text{Adjacent Side to } ∠A } \)

\( \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{ 3}{\sqrt{7}} \)