Differential Calculus-2
Differential Calculus-2
Allen Star Batch
Grade 12
Question:
P and Q are two points on a circle of centre C and radius $a$, the angle PCQ being 20 then the radius of the circle inscribed in the triangle CPQ is maximum when
$\sin \theta = \frac{\sqrt{3}-1}{2\sqrt{2}}$
$\sin \theta = \frac{\sqrt{5}-1}{2}$
$\sin \theta = \frac{\sqrt{5}+1}{2}$
$\sin \theta = \frac{\sqrt{5}-1}{4}$
Step-by-Step Solution
Key Concept: Optimize the inradius by finding critical points of $f(\theta) = \frac{\sin 2\theta}{1+\sin\theta}$ using calculus.
The inradius is given by $r = \frac{\Delta}{s}$ where $\Delta$ is the area and $s$ is the semiperimeter. Substituting $\Delta = \frac{\alpha^2 \sin 2\theta}{2}$ and $s = \alpha + 2\alpha\sin\theta$, we get $r = \frac{\alpha \sin 2\theta}{2(1 + \sin\theta)}$. Setting $f(\theta) = \frac{\sin 2\theta}{1 + \sin\theta}$ and finding $f'(\theta) = 0$ yields the equation $\sin^2\theta + \sin\theta - 1 = 0$, which gives $\sin\theta = \frac{\sqrt{5}-1}{2}$ as the only valid solution in the domain.
Correct Answer: 2