If in a A B C, a^2 ^2 A-b^2-c^2=0, then

MCQ

If in a $\triangle A B C, a^2 \cos ^2 A-b^2-c^2=0$, then

A
$\frac{\pi}{4}< A <\frac{\pi}{2}$
✓
$\frac{\pi}{2}< A <\pi$
C
$A=\frac{\pi}{2}$
D
$A<\frac{\pi}{4}$

✓

Answer

Correct option: B.

$\frac{\pi}{2}< A <\pi$

(B) $a^2 \cos ^2 A-b^2-c^2=0$
$\Rightarrow \cos ^2 A=\frac{ b ^2+ c ^2}{ a ^2}$
Since, $\cos ^2 A \leq 1$ i.e., $\cos ^2 A<1$
$\therefore \quad \frac{ b ^2+ c ^2}{ a ^2}<1 \Rightarrow b^2+ c ^2- a ^2<0$
$\therefore \quad \frac{ b ^2+ c ^2- a ^2}{2 bc }<0$ $\ldots .[\because 2 b c>0]$
$\therefore \cos A <0 \Rightarrow A \in\left(\frac{\pi}{2}, \pi\right)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Trigonometric Functions — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions