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Right Triangle Sides in AP: Sines of Acute Angles | JEE

JEE Maths question with a full step-by-step solution.

Question
If the lengths of the sides of a right triangle are in A.P., then the sines of the acute angles are
A35,45\frac35,\frac45correct
B23,13\sqrt{\frac23},\sqrt{\frac13}
C5−12,5+12\sqrt{\frac{\sqrt5-1}{2}},\sqrt{\frac{\sqrt5+1}{2}}
D3−12,3+12\sqrt{\frac{\sqrt3-1}{2}},\sqrt{\frac{\sqrt3+1}{2}}
Solution
Step 1: Take the sides as a−d, a, a+da-d,\ a,\ a+d (a>da>d), the hypotenuse being a+da+d. By Pythagoras,
(a+d)2=(a−d)2+a2.(a+d)^2=(a-d)^2+a^2.
 Step 2: Expand:
a2+2ad+d2=a2−2ad+d2+a2 ⇒ 4ad=a2 ⇒ d=a4.a^2+2ad+d^2=a^2-2ad+d^2+a^2\ \Rightarrow\ 4ad=a^2\ \Rightarrow\ d=\frac{a}{4}.
 Step 3: The sides are 3a4, a, 5a4=3:4:5\dfrac{3a}{4},\ a,\ \dfrac{5a}{4}=3:4:5, so the acute-angle sines are 35\dfrac35 and 45.\dfrac45. Correct answer: (1)
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