In triangle ABC, with usual notations, if cos | vedclass

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

Question

(C) $1$. Given $\cos A = \frac{1}{2}$, so $A = 60^\circ$ or $\frac{\pi}{3}$.$2$. In $	riangle ABC$, $\angle A + \angle B + \angle C = \pi$. Thus, $\a

Vedclass · Accepted Answer

$b = \sqrt{3}, c = 2\sqrt{3}$

Vedclass · Answer

(C) $1$. Given $\cos A = \frac{1}{2}$, so $A = 60^\circ$ or $\frac{\pi}{3}$.$2$. In $	riangle ABC$, $\angle A + \angle B + \angle C = \pi$. Thus, $\angle C = \pi - (\frac{\pi}{3} + \frac{\pi}{6}) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.$3$. Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$4$. $\frac{3}{\sin(\pi/3)} = \frac{b}{\sin(\pi/6)} = \frac{c}{\sin(\pi/2)}$.$5$. $\frac{3}{\sqrt{3}/2} = \frac{b}{1/2} = \frac{c}{1}$.$6$. $2\sqrt{3} = 2b = c$. Therefore, $b = \sqrt{3}$ and $c = 2\sqrt{3}$.

In $\triangle ABC$, with usual notations, if $\cos A = \frac{1}{2}$, $a = 3$, and $\angle B = \frac{\pi}{6}$, then the values of $b$ and $c$ are:

Similar Questions

In a triangle $ABC$ with usual notations,if $a:b:c = 7:8:9$,then $\cos A : \cos B : \cos C =$

If $a = 2, b = 3, c = 5$ in $\Delta ABC$,then $C = $

In a $\Delta ABC$,if $\cos 3A + \cos 3B + \cos 3C = 1$,then:

Let $ABC$ be an acute-angled triangle and let $D$ be the mid-point of $BC$. If $AB = AD$,then $\tan(B) / \tan(C)$ equals

$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module