In a triangle ABC,if A, B, C are in arithmeti | vedclass

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

Question

(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\ci

Vedclass · Accepted Answer

$2 + \sqrt{3}$

Vedclass · Answer

(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.Substituting $B = 60^{\circ}$ into the given equation: $\cos A + \cos 60^{\circ} + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}}$.$\cos A + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}} - \frac{1}{2} = \frac{1 + \sqrt{2} + \sqrt{3} - \sqrt{2}}{2 \sqrt{2}} = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$.Using the sum-to-product formula: $2 \cos \left( \frac{A+C}{2} ight) \cos \left( \frac{A-C}{2} ight) = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$.Since $A+C = 120^{\circ}$,$\frac{A+C}{2} = 60^{\circ}$,so $2 \cos 60^{\circ} \cos \left( \frac{A-C}{2} ight) = \cos \left( \frac{A-C}{2} ight) = \frac{1 + \sqrt{3}}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}} = \cos 75^{\circ}$.Thus,$\frac{A-C}{2} = 15^{\circ}$,which implies $A-C = 30^{\circ}$.Solving $A+C = 120^{\circ}$ and $A-C = 30^{\circ}$ gives $2A = 150^{\circ}$,so $A = 75^{\circ}$.Finally,$	an A = 	an 75^{\circ} = 	an(45^{\circ} + 30^{\circ}) = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$.

In a triangle $ABC$,if $A, B, C$ are in arithmetic progression and $\cos A + \cos B + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}}$,then $\tan A =$

Similar Questions

In $\triangle ABC$, if $R = \frac{65}{8}$, $rr_1 = 42$ and $r_1 - r = 6.5$, then $s(s-a) = $

If the equation $2 \sin^2 x + \frac{\sin 2x}{2} = k$ has at least one real solution,then the sum of all integral values of $k$ is

In $\triangle ABC$,if $a=2$,$B=\tan ^{-1} \frac{1}{2}$ and $C=\tan ^{-1} \frac{1}{3}$,then $(A, b)$ equals

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

In a $\triangle ABC$,if $\tan A : \tan B : \tan C = 1 : 2 : 3$ and $\sin A : \sin B : \sin C = \sqrt{5} : 2\sqrt{2} : k$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module