If ABC is a triangle of area Delta with a = 2 | vedclass

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

Question

(B) Step $1$: Simplify the expression $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A} = \frac{2 \sin A - 2 \sin A \cos A}{2 \sin A + 2 \sin A \cos A} =

Vedclass · Accepted Answer

$(\frac{3}{4\Delta})^2$

Vedclass · Answer

(B) Step $1$: Simplify the expression $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A} = \frac{2 \sin A - 2 \sin A \cos A}{2 \sin A + 2 \sin A \cos A} = \frac{2 \sin A (1 - \cos A)}{2 \sin A (1 + \cos A)} = \frac{1 - \cos A}{1 + \cos A} = 	an^2(\frac{A}{2})$.Step $2$: Use the half-angle formula $	an^2(\frac{A}{2}) = \frac{(s-b)(s-c)}{s(s-a)}$, where $s = \frac{a+b+c}{2}$.Step $3$: Calculate $s = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$.Step $4$: Substitute values: $	an^2(\frac{A}{2}) = \frac{(4 - 3.5)(4 - 2.5)}{4(4 - 2)} = \frac{0.5 	imes 1.5}{4 	imes 2} = \frac{0.75}{8} = \frac{3}{32}$.Step $5$: Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{4(4-2)(4-3.5)(4-2.5)} = \sqrt{4 	imes 2 	imes 0.5 	imes 1.5} = \sqrt{6}$.Step $6$: Check options: $(\frac{3}{4\Delta})^2 = \frac{9}{16 	imes 6} = \frac{9}{96} = \frac{3}{32}$. Thus, the answer is $(\frac{3}{4\Delta})^2$.

If $ABC$ is a triangle of area $\Delta$ with $a = 2$, $b = \frac{7}{2}$, $c = \frac{5}{2}$, where $a, b, c$ are the lengths of the sides of the triangle opposite to angles $A, B$ and $C$ respectively, then $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A}$ is equal to...

Similar Questions

In $\Delta ABC$,the expression $a(\cos^2 B + \cos^2 C) + \cos A(c \cos C + b \cos B)$ is equal to:

In a triangle $ABC$,if $\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}}$,$a = 2$,$c = 5$ and $b$ is an integer,then the area (in sq. units) of triangle $ABC$ is

In a $\Delta ABC$,if $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$,then the triangle is:

The common solution set of the equations $2 \sin^2 x + \sin^2 2x = 2$ and $\sin 2x + \cos 2x = \tan x$ is

In $\triangle ABC$, $A, B$ and $C$ are in arithmetic progression and $a: c = 1: 2$. If $b = 4 \sqrt{3} \text{ cm}$, then the area of $\triangle ABC$ (in $\text{sq. cm}$) is (in $\sqrt{3}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module