In a triangle with one angle of 120^circ,the | vedclass

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

Question

(B) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a < b < c$. Given $c = 7 \ cm$.Since they are in $A.P.$,$2b = a + c$,so $a = 2b -

Vedclass · Accepted Answer

$\frac{15\sqrt{3}}{4} \ cm^2$

Vedclass · Answer

(B) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a Since they are in $A.P.$,$2b = a + c$,so $a = 2b - 7$.The angle opposite to the greatest side $c$ is $120^\circ$. Using the Law of Cosines:$\cos(120^\circ) = \frac{a^2 + b^2 - c^2}{2ab}$$-\frac{1}{2} = \frac{(2b-7)^2 + b^2 - 7^2}{2(2b-7)b}$$-b(2b-7) = 4b^2 - 28b + 49 + b^2 - 49$$-2b^2 + 7b = 5b^2 - 28b$$7b^2 - 35b = 0$Since $b 
eq 0$,$b = 5 \ cm$.Then $a = 2(5) - 7 = 3 \ cm$.The area of the triangle is $\frac{1}{2}ab \sin(120^\circ) = \frac{1}{2} 	imes 3 	imes 5 	imes \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4} \ cm^2$.

In a triangle with one angle of $120^\circ$,the lengths of the sides form an $A.P.$ If the length of the greatest side is $7 \ cm$,the area of the triangle is:

Similar Questions

If $\alpha$ and $\beta$ are different values of $x$ satisfying $a \cos x + b \sin x = c,$ then $\tan \left( \frac{\alpha + \beta}{2} \right) = $

In $\triangle ABC$,if $\angle C = \frac{\pi}{3}$,then $\frac{3}{a+b+c} - \frac{1}{a+c}$ equals

Let $S = \{\theta \in (0, 2\pi) : 7 \cos^2 \theta - 3 \sin^2 \theta - 2 \cos^2 2\theta = 2\}$. Then,the sum of roots of all the equations $x^2 - 2(\tan^2 \theta + \cot^2 \theta)x + 6 \sin^2 \theta = 0$ for $\theta \in S$ is...

In a $\triangle ABC$,if $\sin^2 B = \sin A$ and $2 \cos^2 A = 3 \cos^2 B$,then the triangle is

The greatest angle of the triangle whose sides are $x^2+x+1$,$2x+1$,and $x^2-1$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module