Question

if A and B are acute angles satisfying sinA=sinB*sinB and 2cosAcosA=3cosBcosB then A+B=

Answer 1

Answer on Question #49554 – Math – Trigonometry

If A and B are acute angles satisfying $\sin A = \sin B \sin B$ and $2\cos A\cos A = 3\cos B\cos B$ then A+B=

Solution

$$\sin B \sin B + \cos B \cos B = 1 = \sin A + \frac{2}{3} \cos A \cos A = \sin A + \frac{2}{3} (1 - \sin^2 A).$$ $$\frac{2}{3} \sin^2 A - \sin A + \frac{1}{3} = 0 \rightarrow \sin^2 A - \frac{3}{2} \sin A + \frac{1}{2} = 0.$$ $$D = \left(- \frac{3}{2}\right)^2 - 4 \cdot \frac{1}{2} = \frac{1}{4}.$$ $$\sin A = \frac{3}{2} \pm \frac{1}{2} = 1; \frac{1}{2}.$$

But A is acute angles, so $\sin A \neq 1$ ($A \neq 90{}^\circ$).

Thus,

$$\sin A = \frac{1}{2} \text{ and } A = 30{}^\circ.$$ $$\sin A = \sin B \sin B = \frac{1}{2} \rightarrow \sin B = \frac{1}{\sqrt{2}} \text{ (angle B is acute too) } \rightarrow B = 45{}^\circ.$$

So,

$$A + B = 75{}^\circ.$$

Answer: $75{}^\circ$.

www.AssignmentExpert.com