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(C) Let the largest angle be $4x$ and the second largest angle be $3x$.Given that the smallest angle is half the largest angle,so the smallest angle $

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Let the largest angle be $4x$ and the second largest angle be $3x$.Given that the smallest angle is half the largest angle,so the smallest angle $= \frac{1}{2} 	imes 4x = 2x$.The sum of the angles in a triangle is $180^{\circ}$.Therefore,$4x + 3x + 2x = 180^{\circ}$.$9x = 180^{\circ}$.$x = 20^{\circ}$.The largest angle $= 4x = 4 	imes 20^{\circ} = 80^{\circ}$.The smallest angle $= 2x = 2 	imes 20^{\circ} = 40^{\circ}$.The difference between the largest and the smallest angle $= 80^{\circ} - 40^{\circ} = 40^{\circ}$.

The largest and the second largest angles of a triangle are in the ratio of $4:3$. The smallest angle is half the largest angle. What is the difference between the smallest and the largest angles of the triangle? (in $^o$)

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