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 the area of a triangle $ABC$ is $\Delta$, then ${a^2}\sin 2B + {b^2}\sin 2A$ is equal to (in $\Delta$)

If the reciprocals of the lengths of the sides of a $\triangle ABC$ are in harmonic progression,then its ex-radii $r_1, r_2, r_3$ are in

If $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$,then a value of $\cos \left(\theta-\frac{\pi}{4}\right)$ among the following is

In a $\Delta ABC$,$a = a_1 = 2$,$b = a_2$,$c = a_3$ such that $a_{p+1} = \frac{5^p}{3^{2-p}} a_p \left( 2^{2-p} - \frac{4p-2}{5^p} a_p \right)$ where $p = 1, 2$,then:

If $x, y, z$ are the lengths of the perpendiculars drawn from the circumcenter to the sides $a, b, c$ respectively of a triangle,then the value of $\frac{bx}{c} + \frac{cy}{a} + \frac{az}{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module