In triangle ABC, if b=6, c=7 and tan frac{A}{ | vedclass

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

Question

(D) Given $b=6, c=7$ and $	an \frac{A}{2}=\frac{1}{\sqrt{6}}$.Using the formula $\cos A = \frac{1-	an^2(A/2)}{1+	an^2(A/2)}$, we get:$\cos A = \fra

Vedclass · Accepted Answer

$\frac{2 \sqrt{6}}{3}$

Vedclass · Answer

(D) Given $b=6, c=7$ and $	an \frac{A}{2}=\frac{1}{\sqrt{6}}$.Using the formula $\cos A = \frac{1-	an^2(A/2)}{1+	an^2(A/2)}$, we get:$\cos A = \frac{1-1/6}{1+1/6} = \frac{5/6}{7/6} = \frac{5}{7}$.Using the Cosine Rule $\cos A = \frac{b^2+c^2-a^2}{2bc}$:$\frac{5}{7} = \frac{6^2+7^2-a^2}{2 	imes 6 	imes 7} = \frac{36+49-a^2}{84}$.$60 = 85 - a^2$ $\Rightarrow a^2 = 25$ $\Rightarrow a=5$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{5+6+7}{2} = 9$.The inradius $r$ is given by $r = (s-a) 	an \frac{A}{2}$.$r = (9-5) 	imes \frac{1}{\sqrt{6}} = \frac{4}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$.

In triangle $ABC$, if $b=6, c=7$ and $\tan \frac{A}{2}=\frac{1}{\sqrt{6}}$, then the inradius of $\triangle ABC$ is

Similar Questions

If $\triangle ABC$ is a right-angled isosceles triangle and $\angle C = 90^{\circ}$,then $r : r_3 =$

$ABC$ is an isosceles triangle with an inscribed circle with center $O$. Let $P$ be the midpoint of $BC$. If $AB = AC = 15$ and $BC = 10$, then $OP$ equals

In a triangle $ABC$, $(r_2+r_3) \operatorname{cosec}^2\left(\frac{A}{2}\right) =$

In a triangle $ABC$, if $a:b:c = 4:5:6$, then $\frac{1}{4R}[r_1+r_2+r_3] =$

In $\triangle ABC$, if $a=7, b=8$ and $c=9$, then $\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module