If in a triangle ABC, $\sin A=sin^2B$ and $2\cos^2A= 3cos ^2B$, then the ΔABC is :
Answer & explanation
Correct answer: option 2
$2(1-\sin^2A)=3(1-\sin^2B);2-2\sin^2A=3-3\sin A$ $[∵\sin^2B=\sin A]$
or $2\sin^2A-3\sin A+1=0$ $∴ (2\sin A -1)(\sin A -1) = 0$
$\sin A=\frac{1}{2}$ or $\sin A = 1$ [Rejected]
As in that case $\sin B =±1 = B = 90°$ and there cannot be two rt. Angles in a triangle
∴ A = 30° or 150° and $\sin^2B=\frac{1}{2}⇒\sin^2B=±\frac{1}{\sqrt{2}}$ ⇒ B = 45° or 135°
⇒ A = 30°, B =45°
⇒ C = 105°
or A = 30°, B = 135°, C = 15°
∴ obtuse; A = 150 (rejected) as in that case A + B alone will be greater than 180°.