Question

If the sides of a right-angle triangle form an A.P., the `Sin' of the acute angles are

• A. \(\left(\frac{3}{5},\frac{4}{5}\right)\)
• B. \(\left(\sqrt{3},\frac{1}{\sqrt{3}}\right)\)
• C. \(\left(\sqrt{\frac{\sqrt5-1}{2}},\sqrt{\frac{\sqrt5-1}{2}}\right)\)
• D. \(\left(\sqrt{\frac{\sqrt3-1}{2}},\sqrt{\frac{\sqrt3-1}{2}}\right)\)

Hint:

Right triangle with sides in A.P. always forms the Pythagorean triple \(3:4:5\).

Answer 1

The Correct Option is A

Step 1: Let sides in A.P. be \(a-d,\ a,\ a+d\).
Since it is a right triangle, largest side is hypotenuse:
\[ a+d \]
Step 2: Apply Pythagoras theorem.
\[ (a-d)^2+a^2=(a+d)^2 \]
Step 3: Expand.
\[ (a^2-2ad+d^2)+a^2=a^2+2ad+d^2 \]
\[ 2a^2-2ad+d^2=a^2+2ad+d^2 \]
\[ a^2=4ad \Rightarrow a=4d \]
Step 4: Find sides ratio.
\[ a-d=4d-d=3d,\quad a=4d,\quad a+d=5d \]
So sides are in ratio:
\[ 3:4:5 \]
Step 5: Sine of acute angles.
\[ \sin\theta=\frac{3}{5},\quad \sin\phi=\frac{4}{5} \]
Final Answer:
\[ \boxed{\left(\frac{3}{5},\frac{4}{5}\right)} \]