If the sides of a right angled triangle form an A.P ,then the sine of the acute angles are
- $\displaystyle \frac{3}{5},\displaystyle \frac{4}{5}$
- $\sqrt {3},\displaystyle \frac{1}{3}$
- $\sqrt{\displaystyle \frac{\sqrt{5}-1}{2}},\sqrt{\displaystyle \frac{\sqrt{5}+1}{2}}$
- $\displaystyle \frac{\sqrt {3}}{2},\displaystyle \frac{1}{2}$
Reveal answer
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A
Correct answer
Explanation
If sides are in A.P., they are a-d, a, a+d. By Pythagorean theorem, (a-d)^2 + a^2 = (a+d)^2. This simplifies to a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2, so a^2 = 4ad, or a = 4d. Sides are 3d, 4d, 5d. Sine of angles are 3/5 and 4/5.
AI explanation
Let the sides of the right angled triangle be a minus d, a, and a plus d, where a plus d is the hypotenuse. Using the Pythagorean theorem, (a - d)^2 + a^2 = (a + d)^2, which simplifies to a = 4d, making the sides 3d, 4d, and 5d. The sine values for the acute angles are the ratios of the opposite sides to the hypotenuse, giving 3d/5d and 4d/5d. The result is 3/5 and 4/5.