Question

If the length of a right triangle are in A.P., then the sines of the acute angles are

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

Answer 1

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

Let the sides be $a-d$, $a$ and $a+d$

Then, by Pyaragoras Theorem,

$(a+d{)}^2=a^2+(a-d{)}^2$

$\Rightarrow 4ad=a^2$

$\Rightarrow a=4d$

therefore sides of triangle are $3d$, $4d$, $5d$. sine of acute angles

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

i.e. $\frac{3}{5},\frac{4}{5}$

Therefore, Option A is correct