In Delta ABC,if cos frac{A}{2} = sqrt{frac{b | vedclass

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

Question

(A) We know the formula for $\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$.Given $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,we equate the two:$\sqr

Vedclass · Accepted Answer

$a^2 + b^2 = c^2$

Vedclass · Answer

(A) We know the formula for $\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$.Given $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,we equate the two:$\sqrt{\frac{s(s - a)}{bc}} = \sqrt{\frac{b + c}{2c}}$Squaring both sides:$\frac{s(s - a)}{bc} = \frac{b + c}{2c}$$\frac{s(s - a)}{b} = \frac{b + c}{2}$$2s(s - a) = b(b + c)$Substitute $2s = a + b + c$ and $s - a = \frac{b + c - a}{2}$:$(a + b + c) \left( \frac{b + c - a}{2} \right) = b^2 + bc$$((b + c) + a)((b + c) - a) = 2(b^2 + bc)$$(b + c)^2 - a^2 = 2b^2 + 2bc$$b^2 + c^2 + 2bc - a^2 = 2b^2 + 2bc$$c^2 - a^2 = b^2$$c^2 = a^2 + b^2$.

In $\Delta ABC$,if $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,then:

Similar Questions

In a triangle $ABC$,if $a = 2$,$B = 60^\circ$ and $C = 75^\circ$,then $b =$

In $\triangle ABC$,if $\frac{1}{r_1}, \frac{1}{r_2}$ and $\frac{1}{r_3}$ are in arithmetic progression,then $r_2 : r =$

In $\Delta ABC$,$(b - c)\cot \frac{A}{2} + (c - a)\cot \frac{B}{2} + (a - b)\cot \frac{C}{2}$ is equal to

If the sides of a triangle are in the ratio $2 : \sqrt{6} : (\sqrt{3} + 1)$,then the largest angle of the triangle will be.....$^o$

In $\Delta ABC$,$({b^2} - {c^2})\cot A + ({c^2} - {a^2})\cot B + ({a^2} - {b^2})\cot C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module