First slide

Arithmetic progression

Question

If the sides of a right-angled triangle are in AP, then the sines of the acute angles are

Easy

A

$\frac{3}{5}, \frac{4}{5}$
B

$\sqrt{3}, \frac{1}{\sqrt{3}}$
C

$\sqrt{\frac{\sqrt{5} - 1}{2}}, \sqrt{\frac{\sqrt{5} + 1}{2}}$
D

$\sqrt{\frac{\sqrt{3} - 1}{2}}, \sqrt{\frac{\sqrt{3} + 1}{2}}$

Solution

Let ABC be the right triangle, right angled at B. Let the sides of the triangle be a-d, a and a + d, with a > d >0.
Clearly, AC =a + d is the largest side.
So, a + d is the side opposite to the right angle.
By Pythagoras theorem, we have
$(a + d)^{2} = a^{2} + (a - d)^{2} \Rightarrow 4 ad = a^{2} \Rightarrow a = 4 d$
Now, $\sin A = \frac{BC}{AC} = \frac{a}{a + d} = \frac{4 d}{5 d} = \frac{4}{5}$
and $\sin C = \frac{AB}{AC} = \frac{a - d}{a + d} = \frac{4 d - d}{4 d + d} = \frac{3}{5}$

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