Q53 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
Let $P_1$, $P_2$, $P_3$ be the altitudes of a triangle ABC from the vertices A, B, C respectively. If $\triangle$ denotes the area of the triangle and s is the semi-perimeter of the triangle, then $\dfrac{\cos A}{P_1} + \dfrac{\cos B}{P_2} + \dfrac{\cos C}{P_3} =$
- A$R$
- Correct answerB$\dfrac{1}{R}$
- C$R^2$
- D$\dfrac{1}{R^2}$
Solution
The area of the triangle is given by $\Delta = \dfrac{1}{2} a P_1 = \dfrac{1}{2} b P_2 = \dfrac{1}{2} c P_3$.
This gives $P_1 = \dfrac{2\Delta}{a}$, $P_2 = \dfrac{2\Delta}{b}$, and $P_3 = \dfrac{2\Delta}{c}$.
Substituting these into the given expression:
$\dfrac{\cos A}{P_1} + \dfrac{\cos B}{P_2} + \dfrac{\cos C}{P_3} = \dfrac{a \cos A + b \cos B + c \cos C}{2\Delta}$
Using the sine rule $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$, the numerator becomes:
$a \cos A + b \cos B + c \cos C = 2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C$
$= R(\sin 2A + \sin 2B + \sin 2C)$
In any triangle $ABC$, the sum of sines of double angles is $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$.
Thus, the numerator is $4R \sin A \sin B \sin C$.
The area of the triangle can be written as $\Delta = \dfrac{abc}{4R} = \dfrac{(2R \sin A)(2R \sin B)(2R \sin C)}{4R} = 2R^2 \sin A \sin B \sin C$.
The denominator is $2\Delta = 4R^2 \sin A \sin B \sin C$.
Dividing the numerator by the denominator:
$\dfrac{4R \sin A \sin B \sin C}{4R^2 \sin A \sin B \sin C} = \dfrac{1}{R}$
Answer: $\dfrac{1}{R}$
This gives $P_1 = \dfrac{2\Delta}{a}$, $P_2 = \dfrac{2\Delta}{b}$, and $P_3 = \dfrac{2\Delta}{c}$.
Substituting these into the given expression:
$\dfrac{\cos A}{P_1} + \dfrac{\cos B}{P_2} + \dfrac{\cos C}{P_3} = \dfrac{a \cos A + b \cos B + c \cos C}{2\Delta}$
Using the sine rule $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$, the numerator becomes:
$a \cos A + b \cos B + c \cos C = 2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C$
$= R(\sin 2A + \sin 2B + \sin 2C)$
In any triangle $ABC$, the sum of sines of double angles is $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$.
Thus, the numerator is $4R \sin A \sin B \sin C$.
The area of the triangle can be written as $\Delta = \dfrac{abc}{4R} = \dfrac{(2R \sin A)(2R \sin B)(2R \sin C)}{4R} = 2R^2 \sin A \sin B \sin C$.
The denominator is $2\Delta = 4R^2 \sin A \sin B \sin C$.
Dividing the numerator by the denominator:
$\dfrac{4R \sin A \sin B \sin C}{4R^2 \sin A \sin B \sin C} = \dfrac{1}{R}$
Answer: $\dfrac{1}{R}$
Hence, the correct answer is option: 'B'.