In triangle ABC with usual notations b=sqrt{3 | vedclass

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

Question

(D) By the cosine rule,we have: $a^2 = b^2 + c^2 - 2bc \cos A$ $a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos(30^{\circ})$ $a^2 = 3 + 1 - 2\sqrt{3}

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) By the cosine rule,we have: $a^2 = b^2 + c^2 - 2bc \cos A$ $a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos(30^{\circ})$ $a^2 = 3 + 1 - 2\sqrt{3} \left(\frac{\sqrt{3}}{2}ight)$ $a^2 = 4 - 3 = 1$ $	herefore a = 1$ Since $b = \sqrt{3} \approx 1.732$ is the largest side,the largest angle is $\angle B$. Using the cosine rule for $\angle B$: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2(1)(1)} = \frac{1 + 1 - 3}{2} = -\frac{1}{2}$ $	herefore B = 120^{\circ}$

In triangle $ABC$ with usual notations $b=\sqrt{3}$,$c=1$,and $m \angle A=30^{\circ}$,then the largest angle of the triangle is (in $^{\circ}$)

Similar Questions

In a triangle $ABC$, if $\sin A \sin B = \frac{ab}{c^2}$, then the triangle is

In a $\triangle ABC$,$2(bc \cos A + ca \cos B + ab \cos C) = ?$

In a $\Delta ABC$,$a = 5, b = 4$ and $\cos(A - B) = \frac{31}{32}$,then side $c$ is equal to

In any triangle,if the angles are in the ratio $1: 2: 3$,then their corresponding sides are in the ratio.

In $\triangle ABC$,if $b+c : c+a : a+b = 7 : 8 : 9$,then the smallest angle (in radians) of that triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module