Q67 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
With the usual notations, if the lengths of the sides of the triangle are 3 units, 5 units and 7 units, then the largest angle of the triangle is
- A$\dfrac{\pi}{2}$
- B$\dfrac{\pi}{3}$
- Correct answerC$\dfrac{2\pi}{3}$
- D$\dfrac{\pi}{4}$
Solution
Let the sides of the triangle be $a = 3$, $b = 5$, and $c = 7$.
The largest angle of a triangle is opposite to its longest side. Thus, the largest angle is $C$, which is opposite to the side $c = 7$.
Using the cosine rule:
$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$
Substituting the given values:
$\cos C = \dfrac{3^2 + 5^2 - 7^2}{2(3)(5)}$
$\cos C = \dfrac{9 + 25 - 49}{30}$
$\cos C = \dfrac{34 - 49}{30}$
$\cos C = \dfrac{-15}{30} = -\dfrac{1}{2}$
Since $0 < C < \pi$, we get $C = \dfrac{2\pi}{3}$.
The largest angle of a triangle is opposite to its longest side. Thus, the largest angle is $C$, which is opposite to the side $c = 7$.
Using the cosine rule:
$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$
Substituting the given values:
$\cos C = \dfrac{3^2 + 5^2 - 7^2}{2(3)(5)}$
$\cos C = \dfrac{9 + 25 - 49}{30}$
$\cos C = \dfrac{34 - 49}{30}$
$\cos C = \dfrac{-15}{30} = -\dfrac{1}{2}$
Since $0 < C < \pi$, we get $C = \dfrac{2\pi}{3}$.
Hence, the correct answer is option: 'C'.