If in a $Δ A B C, a^{2} {cos}^{2} A = b^{2} + c^{2}$ then

A < \frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4} < A < \frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A > \frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A = \frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $A > \frac{π}{2}$
Given, $a^{2} {cos}^{2} A = b^{2} + c^{2}$
$\Rightarrow a^{2} (1 - {sin}^{2} A) = b^{2} + c^{2}$
$\Rightarrow a^{2} {sin}^{2} A = - (b^{2} + c^{2} - a^{2})$
Now using cosine rule $cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{- a^{2} {sin}^{2} A}{2 b c} < 0$
$∴ A > \frac{π}{2}$

Suggest Corrections

Similar questions

If in a $△ A B C, a^{2} c o s^{2} A = b^{2} + c^{2}$ then

z,

| z - 1 + i | + | z + i | = 1,

z

(

\in (- π, π])

π

s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} = 1 + 4 s i n [\frac{π - A}{4}] s i n [\frac{π - B}{4}] s i n [\frac{π - C}{4}]

tan x > cot x

x \in (0, π) - {\frac{π}{2}}

Δ A B C

a = 5; b = 4, A = \frac{π}{2} + B

tan C =

Join BYJU'S Learning Program