If two sides of a triangle are sqrt{3}-2 and | vedclass

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

Question

(D) Let the sides of the triangle be $a = \sqrt{3}-2$ and $b = \sqrt{3}+2$,and the included angle be $C = 60^{\circ}$.Using the Law of Cosines,the thi

Vedclass · Accepted Answer

$\sqrt{15}$ units

Vedclass · Answer

(D) Let the sides of the triangle be $a = \sqrt{3}-2$ and $b = \sqrt{3}+2$,and the included angle be $C = 60^{\circ}$.Using the Law of Cosines,the third side $c$ is given by:$c^2 = a^2 + b^2 - 2ab \cos(C)$$c^2 = (\sqrt{3}-2)^2 + (\sqrt{3}+2)^2 - 2(\sqrt{3}-2)(\sqrt{3}+2) \cos(60^{\circ})$$c^2 = (3 + 4 - 4\sqrt{3}) + (3 + 4 + 4\sqrt{3}) - 2(3 - 4) 	imes \frac{1}{2}$$c^2 = 7 - 4\sqrt{3} + 7 + 4\sqrt{3} - 2(-1) 	imes \frac{1}{2}$$c^2 = 14 + 1 = 15$$c = \sqrt{15}$ units.

If two sides of a triangle are $\sqrt{3}-2$ and $\sqrt{3}+2$ units and their included angle is $60^{\circ}$,then the third side of the triangle is

Similar Questions

In a $\triangle ABC$,if $4a = b + c$,then $\tan \frac{B}{2} \tan \frac{C}{2} =$

In $\triangle ABC$,if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$,then $\tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{C}{2}\right) = $

If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

Let $\Delta$ denote the area of a $\triangle ABC$. If $\alpha, \beta, \gamma$ are the lengths of the altitudes of the $\triangle ABC$,then $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$

If $b+c=3a$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module