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
$\text{AC} ^2 = \text{AB} ^2 + \text{BC}^ 2$
$(4k) ^2 = \text{AB}^ 2 + (3k)^ 2$
$16k ^2 - 9k^ 2 =\text{ AB}^ 2$
$7k ^2 = \text{AB}^ 2$
$AB =\sqrt7k.$

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

$\frac{AB}{AC} =\frac{ \sqrt7k}{4k} =\frac{ \sqrt7}{4}$

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

$\frac{BC}{AB} = \frac{3k}{\sqrt7k} =\frac{ 3}{\sqrt7}$