The circular measure of an angle of an isosce | vedclass

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(C) The sum of all angles in a triangle is $\pi$ radians.Given one angle is $\frac{5 \pi}{9}$.The sum of the remaining two angles is $\pi - \frac{5 \p

Vedclass · Accepted Answer

$\frac{2 \pi}{9}$

Vedclass · Answer

(C) The sum of all angles in a triangle is $\pi$ radians.Given one angle is $\frac{5 \pi}{9}$.The sum of the remaining two angles is $\pi - \frac{5 \pi}{9} = \frac{4 \pi}{9}$.Since the triangle is isosceles,there are two possible cases:Case $1$: The given angle is the vertex angle. Then the two base angles are equal. Each base angle $= \frac{1}{2} 	imes \frac{4 \pi}{9} = \frac{2 \pi}{9}$.Case $2$: The given angle is one of the base angles. Then the other base angle is also $\frac{5 \pi}{9}$. However,the sum of two angles $\frac{5 \pi}{9} + \frac{5 \pi}{9} = \frac{10 \pi}{9}$,which is greater than $\pi$. This is impossible for a triangle.Therefore,the other angles must be $\frac{2 \pi}{9}$.

The circular measure of an angle of an isosceles triangle is $\frac{5 \pi}{9}$. The circular measure of one of the other angles must be

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