Question

If the sides of a right angled triangle form an A.P ,then the sine of the acute angles are

• A. $\frac{3}{5},\frac{4}{5}$
• B. $\sqrt{3},\frac{1}{3}$
• C. $\sqrt{\frac{\sqrt{5}-1}{2}},\sqrt{\frac{\sqrt{5}+1}{2}}$
• D. $\frac{\sqrt{3}}{2},\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{5},\frac{4}{5}$

Let sides be $a-d,a,a+d$ in A.P.
$\Rightarrow {(a+d)}^2={(a-d)}^2+a^2$
$\Rightarrow 4ad=a^2$
$\Rightarrow a=4d$
Therefore, the sides are $3d,4d$ and $5d.$
Consider the triangle ABC with $\angle B={90}^{\circ }$.
Let $AB=a,BC=a-d,AC=a+d$ be the sides of the triangle.
Therefore, $\sin A=\frac{a-d}{a+d}$
$\Rightarrow \sin A=\frac{3d}{5d}$
$\Rightarrow \sin A=\frac{3}{5}$
Also, $\sin C=\frac{a}{a+d}$
$\Rightarrow \sin C=\frac{4d}{5d}$
$\Rightarrow \sin C=\frac{4}{5}$
Hence, option 'A' is correct.