In Delta ABC,if a^{2} cos^{2} A - b^{2} - c^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,$a^{2} \cos^{2} A - b^{2} - c^{2} = 0$ $\Rightarrow a^{2} \cos^{2} A = b^{2} + c^{2}$ Using the Law of Cosines,$\cos A = \frac{b^{2} + c^{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given,$a^{2} \cos^{2} A - b^{2} - c^{2} = 0$ $\Rightarrow a^{2} \cos^{2} A = b^{2} + c^{2}$ Using the Law of Cosines,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$. Substituting $b^{2} + c^{2} = a^{2} \cos^{2} A$,we get: $\cos A = \frac{a^{2} \cos^{2} A - a^{2}}{2bc} = \frac{-a^{2}(1 - \cos^{2} A)}{2bc} = \frac{-a^{2} \sin^{2} A}{2bc}$. Since $a, b, c > 0$ and $\sin^{2} A > 0$ for $0 Therefore,$A$ must lie in the second quadrant,i.e.,$\frac{\pi}{2} < A < \pi$.

In $\Delta ABC$,if $a^{2} \cos^{2} A - b^{2} - c^{2} = 0$,then

Similar Questions

If $a, b, c$ are the sides and $A, B, C$ are the angles of a triangle $ABC$,then $\tan \left( \frac{A}{2} \right)$ is equal to

In a triangle $ABC$,if $(b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right) = K(1 - \cos 2A)$,then $K =$

In a triangle $ABC$ with usual notations,$\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}=$

If in a $\triangle ABC$,$r_1 < r_2 < r_3$,then:

$A$ triangle can be uniquely determined by its

Vedclass Test Series

Exam Paper Generator

Online Exam Module