Question

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

Answer 1

Answer:

11/9

Step-by-step explanation:

3cosA-4sinA=0

cosA=4/3sinA

$\frac{sinA+2cosA}{3cosA-sinA} =\frac{sinA+2 \times \frac{4}{3}sin A}{4sinA-sinA} \\=\frac{sinA(1+\frac{8}{3})}{3 sinA} \\=\frac{3+8}{9} \\=\frac{11}{9}$

Answer 2

$\large\underline{\sf{Given\: info-}}$

3 cos A - 4 sin A = 0

To Evaluate:

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

$\large\underline{\sf{Solution-}}$

$\sf\longmapsto 3\cos A-4\sin A=0$

Transposing 4 sin A to RHS,

$\sf\longmapsto 3\cos A=4\sin A$

Transposing 4 to LHS and cos A to RHS,

$\sf\longmapsto \dfrac{3}{4}=\dfrac{\sin A}{\cos A}$

So,

$\sf\longmapsto \tan A=\dfrac{3}{4}$

Now, as $\sf tan A = \dfrac{3}{4}$,

$\sf\longmapsto \sin A=\dfrac{3}{5}$

And,

$\sf\longmapsto \cos A=\dfrac{4}{5}$

We need to find that,

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

So,

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

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

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

So,

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

So,

$\sf\longmapsto \dfrac{11}{9}$

Answer:-

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