Question

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

Answer 1

Let \(ΔABC\) be a right-angled triangle, right-angled at point B.

Given that, 
sin A=\(\frac{3}{4}\)

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

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. 
Applying Pythagoras theorem in \(ΔABC,\) we obtain 
\(\text{AC} ^2 = \text{AB} ^2 + \text{BC}^ 2\)
\((4k) ^2 = \text{AB}^ 2 + (3k)^ 2\)
\(16k ^2 - 9k^ 2 =\text{ AB}^ 2\)
\(7k ^2 = \text{AB}^ 2\)
\(AB =\sqrt7k.\)

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

\(\frac{AB}{AC} =\frac{ \sqrt7k}{4k} =\frac{ \sqrt7}{4}\)

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

\(\frac{BC}{AB} = \frac{3k}{\sqrt7k} =\frac{ 3}{\sqrt7}\)