If the sum of the length of the hypotenuse and a side of a right angled triangle is L cm then the angle between hypotenuse and other side of the triangle for which the area of the triangle is maximum, is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\pi}{6} \)
Let the given side be x and the hypotenuse h, with x + h = L. The other side is \( \sqrt{h^2-x^2} \), and it is easier to maximise the square of the area:
\( \displaystyle A^2=\frac{1}{4}x^2(h^2-x^2) \)
\( \displaystyle =\frac{1}{4}x^2\left[(L-x)^2-x^2\right] \)
\( \displaystyle =\frac{1}{4}(L^2x^2-2Lx^3) \)
\( \displaystyle \frac{d(A^2)}{dx}=\frac{1}{4}(2L^2x-6Lx^2)=0 \)
\( \displaystyle x=\frac{L}{3} \)
Then \( h=\frac{2L}{3} \). The given side x is opposite the angle \( \theta \) between the hypotenuse and the other side, so
\( \displaystyle \sin\theta=\frac{x}{h}=\frac{1}{2} \)
\( \displaystyle \theta=\frac{\pi}{6} \)
Why the other options are incorrect
Option (2) \( \displaystyle\frac{\pi}{3} \) — Incorrect. This is the angle between the hypotenuse and the given side (its cosine is \( \frac{x}{h}=\frac{1}{2} \)); the question asks for the angle with the other side.
Option (3) \( \displaystyle\frac{\pi}{4} \) — Incorrect. \( \sin\theta=\frac{1}{2} \), not \( \frac{1}{\sqrt{2}} \).
Option (4) \( \displaystyle\frac{\pi}{5} \) — Incorrect. \( \sin\theta=\frac{1}{2} \) gives \( \frac{\pi}{6} \), not \( \frac{\pi}{5} \).