Question

If sinA/3= sinB/4=1/5 and A, B are angles
in the second quadrant then prove that
4cosA + 3cosB = -5.​

Answer 1

Answer:

$4\:cosA+3\:cosB=5$

Step-by-step explanation:

Given:

$\frac{sinA}{3}=\frac{sinB}{4}=\frac{1}{5}$

$\implies\:\frac{sinA}{3}=\frac{1}{5}$

$\implies\:sinA=\frac{3}{5}$

similarly,

$\implies\:sinB=\frac{4}{5}$

Now,

$cos^2A=1-sin^2A$

$cos^2A=1-(\frac{3}{5})^2$

$\implies\:cos^2A=1-\frac{9}{25}$

$\implies\:cos^2A=\frac{25-9}{25}$

$\implies\:cos^2A=\frac{16}{25}$

$\implies\:cosA=\pm\frac{4}{5}$

since A lies in second quadrant,

$cosA=\frac{-4}{5}$

$cos^2B=1-sin^2B$

$cos^2B=1-(\frac{4}{5})^2$

$\implies\:cos^2B=1-\frac{16}{25}$

$\implies\:cos^2B=\frac{25-16}{25}$

$\implies\:cos^2B=\frac{9}{25}$

$\implies\:cosB=\pm\frac{3}{5}$

since B lies in second quadrant,

$cosA=\frac{-3}{5}$

$4\:cosA+3\:cosB$

$=4(\frac{-4}{5})+3(\frac{-3}{5})$

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

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

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

$=-5$

$\implies\:4\:cosA+3\:cosB=5$