In a triangle ABC,a=1,b=sqrt{3} and angle C=p | vedclass

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

Question

(C) In a $	riangle ABC$,it is given that $a=1$,$b=\sqrt{3}$ and $\angle C=\frac{\pi}{6}$.Using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) In a $	riangle ABC$,it is given that $a=1$,$b=\sqrt{3}$ and $\angle C=\frac{\pi}{6}$.Using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$\cos \frac{\pi}{6} = \frac{1^2 + (\sqrt{3})^2 - c^2}{2(1)(\sqrt{3})}$$\frac{\sqrt{3}}{2} = \frac{1 + 3 - c^2}{2\sqrt{3}}$$\frac{\sqrt{3}}{2} = \frac{4 - c^2}{2\sqrt{3}}$Multiply both sides by $2\sqrt{3}$:$3 = 4 - c^2$$c^2 = 4 - 3 = 1$Since $c$ represents the length of a side,$c = 1$.Therefore,option $C$ is correct.

In a $\triangle ABC$,$a=1$,$b=\sqrt{3}$ and $\angle C=\pi/6$. Then the measure of the third side $c=$

Similar Questions

If in a triangle $ABC$,$b = \sqrt{3}$,$c = 1$ and $B - C = 90^{\circ}$,then $\angle A$ is .....$^{\circ}$.

In a $\Delta ABC$,the expression $\left( \frac{a^2}{\sin A} + \frac{b^2}{\sin B} + \frac{c^2}{\sin C} \right) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ simplifies to,where $\Delta$ is the area of the triangle.

In a $\triangle ABC$,if $\tan(A/2)$,$\tan(B/2)$,and $\tan(C/2)$ are in Arithmetic Progression,then which of the following options is always correct?

In a triangle $ABC$,$b^2 \sin 2C + c^2 \sin 2B =$

If the perimeter of the triangle $ABC$ is $50 \text{ cm}$,then $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module