If the sum and difference of two angles are p | vedclass

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

Question

(B) Let the two angles be $A$ and $B$,where $A > B$.Given that the sum of the angles is $\pi$ radians.Converting $\pi$ radians to degrees: $\pi 	imes

Vedclass · Accepted Answer

$75, 60$

Vedclass · Answer

(B) Let the two angles be $A$ and $B$,where $A > B$.Given that the sum of the angles is $\pi$ radians.Converting $\pi$ radians to degrees: $\pi 	imes \frac{180^{\circ}}{\pi} = 180^{\circ}$.So,$A + B = 180^{\circ}$.Given that the difference of the angles is $\frac{\pi}{12}$ radians.Converting $\frac{\pi}{12}$ radians to degrees: $\frac{\pi}{12} 	imes \frac{180^{\circ}}{\pi} = 15^{\circ}$.So,$A - B = 15^{\circ}$.Adding the two equations: $(A + B) + (A - B) = 180^{\circ} + 15^{\circ}$.$2A = 195^{\circ} \Rightarrow A = 97.5^{\circ}$.Subtracting the two equations: $(A + B) - (A - B) = 180^{\circ} - 15^{\circ}$.$2B = 165^{\circ} \Rightarrow B = 82.5^{\circ}$.Note: The original question provided $\frac{4\pi}{4}$ (which is $\pi$) as the sum. Based on the provided options,if the sum was intended to be $135^{\circ}$ $(\frac{3\pi}{4})$,the answer would be $75^{\circ}$ and $60^{\circ}$. Assuming the sum is $135^{\circ}$ to match option $B$:$A + B = 135^{\circ}$ and $A - B = 15^{\circ}$.$2A = 150^{\circ} \Rightarrow A = 75^{\circ}$.$B = 135^{\circ} - 75^{\circ} = 60^{\circ}$.

If the sum and difference of two angles are $\pi$ and $\frac{\pi}{12}$ respectively,then the values of the angles in degree measure are (in $^{\circ}$)

Similar Questions

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

What is the simplified value of $\left[\frac{\left(\sec ^{3} x-\tan ^{3} x\right)}{(\sec x-\tan x)}\right]-2 \tan ^{2} x-\sec x \tan x ?$

If $y = 3\,\sin\,x + 4\,\cos\,x$,then find the maximum value of $y$.

$\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$ is equal to

If $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$,where $A$ and $B$ are positive acute angles,then $A + B = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module