The difference between the two acute angkes of a right angles triangle is 2 5 redians. Express the angles in degrees.

Let _ 1 and _ 2 be two acute angles of a right angles triangle. Difference of acute angles. _ 1 - _ 2 =2 5 radians In a right angled triangle, _ 1 + _ 2 = 2 _ 1 + _ 2 =2 5 _ 1 + _ 2 = 2 On solving 2 _ 1 =2 5 + 2 _ 1 =9 20 From quation (ii) _ 2 = 20 So angles in degrees, _ 1 =9 20 180 =81^ _ 2 = 20 180 =9^

The difference between the two acute angkes of a right angles tri…

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The difference between the two acute angkes of a right angles triangle is $\frac{2\pi}{5}$ redians. Express the angles in degrees.

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Answer

Let $\theta_{1}$ and $\theta_{2}$ be two acute angles of a right angles triangle. $\therefore$ Difference of acute angles. $\theta_{1}-\theta_{2}=\frac{2\pi}{5}\ \text{radians}$ $\therefore$ In a right angled triangle, $\theta_{1}+\theta_{2}=\frac{\pi}{2}$ $\theta_{1}+\theta_{2}=\frac{2\pi}{5}$ $\theta_{1}+\theta_{2}=\frac{\pi}{2}$ On solving $2\theta_{1}=\frac{2\pi}{5}+\frac{\pi}{2}$ $\theta_{1}=\frac{9\pi}{20}$ From quation (ii) $\theta_{2}=\frac{\pi}{20}$ So angles in degrees, $\theta_{1}=\frac{9\pi}{20}\times\frac{180}{\pi}=81^{\circ}$ $\theta_{2}=\frac{\pi}{20}\times\frac{180}{\pi}=9^{\circ}$

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