Question

If the sides of a right angle triangle form an A.P. then the sines of the acute angles are

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

Answer 1

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

Let sides of the triangle are $a,b$ and $\sqrt{a^2+b^2}$

given sides are in A,P so $\Rightarrow$ $2b$ $=$ $a+$$\sqrt{a^2+b^2}$

$\Rightarrow$ $2b$ $-a$ $=$$\sqrt{a^2+b^2}$

squaring both side,

$(2b-a{)}^2$ $=$ $a^2+b^2$ $\Rightarrow$ $4b^2+a^2-4ab$ $=$ $a^2+b^2$ $\Rightarrow$ $3b$ $=$ $4a$

now $\sqrt{a^2+b^2}$ $=$ $\sqrt{a^2+(\frac{4a}{3}{)}^2}$ $=$ $\frac{5a}{3}$

so the sides of the triangle are $a,\frac{4a}{3},\frac{5a}{3}$

Using sine rule, we get the sines of the acute angles as $\frac{4}{5}and\frac{3}{5}$

Hence, option 'A' is correct.