Q66 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
In $\triangle ABC$, with usual notations, if $\cos A = \dfrac{1}{2}, a = 3, \angle B = \dfrac{\pi^c}{6}$, then the values of $b$ and $c$ are.....
- A$b = \sqrt{3}, c = \sqrt{3}$
- B$b = 2\sqrt{3}, c = \sqrt{3}$
- Correct answerC$b = \sqrt{3}, c = 2\sqrt{3}$
- D$b = \sqrt{3}, c = \dfrac{1}{\sqrt{3}}$
Solution
Given $\cos A = \dfrac{1}{2}$, we have $A = \dfrac{\pi}{3}$.
The sum of angles in a triangle is $\pi$, so $C = \pi - (A + B) = \pi - \left(\dfrac{\pi}{3} + \dfrac{\pi}{6}\right) = \dfrac{\pi}{2}$.
Using the sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$.
Substituting the given values, $\dfrac{3}{\sin(\pi/3)} = \dfrac{b}{\sin(\pi/6)} = \dfrac{c}{\sin(\pi/2)}$.
$\dfrac{3}{\dfrac{\sqrt{3}}{2}} = \dfrac{b}{\dfrac{1}{2}} = \dfrac{c}{1}$.
$2\sqrt{3} = 2b = c$.
Therefore, $b = \sqrt{3}$ and $c = 2\sqrt{3}$.
Answer: $b = \sqrt{3}, c = 2\sqrt{3}$
The sum of angles in a triangle is $\pi$, so $C = \pi - (A + B) = \pi - \left(\dfrac{\pi}{3} + \dfrac{\pi}{6}\right) = \dfrac{\pi}{2}$.
Using the sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$.
Substituting the given values, $\dfrac{3}{\sin(\pi/3)} = \dfrac{b}{\sin(\pi/6)} = \dfrac{c}{\sin(\pi/2)}$.
$\dfrac{3}{\dfrac{\sqrt{3}}{2}} = \dfrac{b}{\dfrac{1}{2}} = \dfrac{c}{1}$.
$2\sqrt{3} = 2b = c$.
Therefore, $b = \sqrt{3}$ and $c = 2\sqrt{3}$.
Answer: $b = \sqrt{3}, c = 2\sqrt{3}$
Hence, the correct answer is option: 'C'.