If $\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}$ where $A$ and $B$ are the angles in $2^{n d}$ quadrant, then find the value of

$(4 cos A + 3 cos B)$ .

Open in App

Solution

$\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}$

Angles $A$ & $B$ are in $2^{n d}$ quadrant.

$∴ cos A = - \sqrt{1 - {sin}^{2} A} = - \sqrt{1 - \frac{9}{25}} = - \sqrt{\frac{16}{25}} = \frac{- 4}{5}$

& $cos B = - \sqrt{1 - {sin}^{2} B} = - \sqrt{1 - \frac{16}{25}} = - \sqrt{\frac{9}{25}} = \frac{- 3}{5}$

$∴ 4 cos A + 3 cos B$

$= \frac{- 16}{5} - \frac{9}{5} = \frac{- 25}{5} = - 5$

Suggest Corrections

Similar questions

\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}

4 cos A + 3 cos B

sin A + sin B = \sqrt{3} (cos B - cos A)

sin 3 A + sin 3 B

sin A = \frac{1}{\sqrt{10}}

sin B = \frac{1}{\sqrt{5}}

A

B

A + B =

If sinA = $\frac{1}{\sqrt{10}}$ and sinB = $\frac{1}{\sqrt{5}}$ , where A and B are positive acute angles, then A+B=

s i n B = \frac{1}{2}

c o s B = \frac{\sqrt{3}}{2}

3 c o s B - 4 c o s^{3} B

Join BYJU'S Learning Program