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Sequences & Series: Right Triangle Sides Solving Pythagoras Gives Ratio Acute

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

Question
If a right triangle has sides in AP, solving Pythagoras gives ratio 3:4:5, so the acute angle sines are 3/5 and 4/5
A(0,∞)(0,\infty)
B(1,∞)(1,\infty)
C(3,∞)(3,\infty)correct
D(1,3)(1,3)
Solution
Step 1: Let pp be the sum. By AM–GM,
p3≥(xln⁡y−ln⁡z⋅yln⁡z−ln⁡x⋅zln⁡x−ln⁡y)1/3.\frac{p}{3}\ge\left(x^{\ln y-\ln z}\cdot y^{\ln z-\ln x}\cdot z^{\ln x-\ln y}\right)^{1/3}.
 Step 2: Take ln⁡\ln of the product:
(ln⁡y−ln⁡z)ln⁡x+(ln⁡z−ln⁡x)ln⁡y+(ln⁡x−ln⁡y)ln⁡z=0 ⇒ product=1.(\ln y-\ln z)\ln x+(\ln z-\ln x)\ln y+(\ln x-\ln y)\ln z=0\ \Rightarrow\ \text{product}=1.
 Step 3: So p≥3p\ge 3, strict since x,y,zx,y,z are distinct ⇒p∈(3,∞).\Rightarrow p\in(3,\infty). Correct answer: (3)
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