Question

Given 3 cosA - 4 sinA = 0; evaluate without
sin A +2cos A
using tables:
3 cos A-sin A​

Answer 1

ANSWER:

Given:

• 3 cos A - 4 sin A = 0

To Evaluate:

• (sin A + 2 cos A)/(3 cos A - sin A)

Solution:

We are given that,

$\implies 3\cos A-4\sin A=0$

Transposing 4 sin A to RHS,

$\implies 3\cos A=4\sin A$

Transposing 4 to LHS and cos A to RHS,

$\implies \dfrac{3}{4}=\dfrac{\sin A}{\cos A}$

So,

$\implies \tan A=\dfrac{3}{4}$

Now, as tan A = 3/4,

$\implies \sin A=\dfrac{3}{5}$

And,

$\implies \cos A=\dfrac{4}{5}$

We need to find that,

$\implies \dfrac{\sin A+2\cos A}{3\cos A-\sin A}$

So,

$\implies \dfrac{\left(\frac{3}{5}\right)+2\left(\frac{4}{5}\right)}{3\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right) }$

$\implies \dfrac{\left(\frac{3}{5}\right)+\left(\frac{8}{5}\right)}{\left(\frac{12}{5}\right)-\left(\frac{3}{5}\right) }$

$\implies \dfrac{\left(\frac{3+8}{5}\right)}{\left(\frac{12-3}{5}\right)}$

So,

$\implies \dfrac{\left(\frac{11}{5\!\!\!/}\right)}{\left(\frac{9}{5\!\!\!/}\right)}$

So,

$\implies \dfrac{11}{9}$

Therefore,

$\implies\bf \dfrac{sin A+2cos A}{3cos A-sin A}=\dfrac{11}{9}$