Question

if Cos A equal to 3 / 5, then find the value of Sin A - Cos A / 2 tan A​

Answer 1

Given,

$\cos(A \: ) = \frac{3}{5}$

To find out,

$\frac{ \sin(A \: ) \: - \cos(A \: ) }{2 \: \tan(A) }$

Solution:

We know that,

$\cos(A \: ) = \frac{adjacent \: side \: to \:A \: }{hypotenuse}$

In a right angle triangle,

<A = Ѳ

<B = 90°

<C = 90°- Ѳ

AB = Adjacent side to <A = 3

AC = hypotenuse = 5

BC = Opposite side to <A = ?

By pythagoras theorem,

${AC}^{2} = {AB}^{2} + {BC}^{2}$

${5}^{2} = {3}^{2} + {BC}^{2}$

$25 = 9 + {BC}^{2}$

$25 - 9 = {BC}^{2}$

$16 = {BC}^{2}$

$\sqrt{16} = BC$

$4 = BC$

Therefore the value of BC is 4.

Now,

$\sin(A) = \frac{opposite \: side \: to \: A}{hypotenuse} = \frac{4}{5}$

$\tan(A) = \frac{opposite \: side \: to \:A }{adjecent \: side \: to \:A } = \frac{4}{3}$

Substitute these values in

$\frac{ \sin(A \: ) \: - \cos(A \: ) }{2 \: \tan(A) }$

$\frac{ \frac{4}{5} - \frac{3}{5} }{2 \times \frac{4}{3} }$

$\frac{ \frac{4 - 3}{5} }{ \frac{8}{3} }$

$\frac{ \frac{1}{5} }{ \frac{8}{3} }$

$\frac{1}{5} \times \frac{3}{8}$

$\frac{3}{40}$

Therefore the value is 3/40.