Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade None
Question:
If $P$ be any interior point of the equilateral $\triangle ABC$ of side length $2$ units and also $x_a, x_b, x_c$ be the distances of $P$ from the sides $BC, CA, AB$ respectively, then $x_a + x_b + x_c =$
$\sqrt{3}$
$3\sqrt{2}$
$4$
None of these
Step-by-Step Solution
Key Concept: The area of the original triangle equals the sum of the areas of three sub-triangles formed by connecting the point $P$ to each vertex.
Let $A_1$, $B_1$, $C_1$ be the feet of altitudes from $P$ to sides $BC$, $CA$, $AB$ respectively. Given $PA_1 = x_a$, $PB_1 = x_b$, $PC_1 = x_c$, we have $\triangle ABC = \triangle BPC + \triangle CPA + \triangle APB$. Therefore $\frac{\sqrt{3}}{4} \cdot 4 = \frac{1}{2}(2x_a + 2x_b + 2x_c)$, which simplifies to $x_a + x_b + x_c = \sqrt{3}$.
Correct Answer: 1