In any triangle ABC, the expression frac{(a+b | vedclass

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

Question

(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of $	riangle ABC$. Then $a+b+c = 2s$, $b+c-a = 2(s-a)$, $c+a-b = 2(s-b)$, and $a+b-c = 2(s-c)$.Su

Vedclass · Accepted Answer

$\sin^2 A$

Vedclass · Answer

(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of $	riangle ABC$. Then $a+b+c = 2s$, $b+c-a = 2(s-a)$, $c+a-b = 2(s-b)$, and $a+b-c = 2(s-c)$.Substituting these into the expression:$\frac{(2s)(2(s-a))(2(s-b))(2(s-c))}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2}$.Using Heron's formula, $\Delta^2 = s(s-a)(s-b)(s-c)$, so the expression becomes $\frac{16\Delta^2}{4b^2c^2} = \frac{4\Delta^2}{b^2c^2}$.Since the area $\Delta = \frac{1}{2}bc \sin A$, we have $\sin A = \frac{2\Delta}{bc}$.Thus, $\frac{4\Delta^2}{b^2c^2} = (\frac{2\Delta}{bc})^2 = \sin^2 A$.

In any $\triangle ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

Similar Questions

If $ABC$ is a right-angled triangle with $90^{\circ}$ at $C$ and $a > b$,then $\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = $

The sides of a $\triangle ABC$ are given by $a=3, b=5$ and $c=3$. Then,$\cos A=$

If $R : r_1 : r = 5 : 12 : 2$, then $r + r_3 + r_2 - r_1 = $

Given $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$,then the ordered triple $(\alpha, \beta, \gamma)$ is proportional to

In $\triangle ABC$,if $AB: BC: CA = 6: 4: 5$,then $R: r =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module