Question

If the sides of a right-angled triangle form an A.P. Then the sines of the acute angle are

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

Answer 1

The correct option is A

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

Let ABC be the right triangle, right angled at B.
Let the sides of the triangle be a-d, a, a+d with a>d>0.
Clearly AC =a+d is the largest side so, a+d is the hypotenuse of the triangle.
By Pythagoras theorem we have
$(a+d{)}^2=a^2+(a-d{)}^2$
$\Rightarrow 4ad=a^2$
$\Rightarrow a=4d$
sinA=$\frac{BC}{AC}=\frac{a}{a+d}=\frac{4d}{5d}=\frac{4}{5}$
sinC=$\frac{AB}{BC}=\frac{a-d}{a+d}=\frac{4d-d}{4d+d}=\frac{3}{5}$