Q127 / 240 · KCET 2026 · Mathematics
The angles of a triangle are in A.P and the greatest angle is double the least angle, then sine of the third angle is
- Correct answerA$\dfrac{\sqrt{3}}{2}$
- B$\dfrac{1}{2}$
- C$\dfrac{1}{\sqrt{2}}$
- D$0$
Solution
Let the angles of the triangle be $A-D$, $A$, and $A+D$.
The sum of the angles of a triangle is $180^{\circ}$.
$(A-D) + A + (A+D) = 180^{\circ}$
$3A = 180^{\circ} \Rightarrow A = 60^{\circ}$
The greatest angle is double the least angle.
$A+D = 2(A-D)$
Substituting $A = 60^{\circ}$:
$60^{\circ}+D = 2(60^{\circ}-D)$
$60^{\circ}+D = 120^{\circ}-2D$
$3D = 60^{\circ} \Rightarrow D = 20^{\circ}$
The angles of the triangle are $40^{\circ}$, $60^{\circ}$, and $80^{\circ}$.
The third angle (middle angle) is $60^{\circ}$.
The sine of the third angle is $\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$.
Answer: $\dfrac{\sqrt{3}}{2}$
The sum of the angles of a triangle is $180^{\circ}$.
$(A-D) + A + (A+D) = 180^{\circ}$
$3A = 180^{\circ} \Rightarrow A = 60^{\circ}$
The greatest angle is double the least angle.
$A+D = 2(A-D)$
Substituting $A = 60^{\circ}$:
$60^{\circ}+D = 2(60^{\circ}-D)$
$60^{\circ}+D = 120^{\circ}-2D$
$3D = 60^{\circ} \Rightarrow D = 20^{\circ}$
The angles of the triangle are $40^{\circ}$, $60^{\circ}$, and $80^{\circ}$.
The third angle (middle angle) is $60^{\circ}$.
The sine of the third angle is $\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$.
Answer: $\dfrac{\sqrt{3}}{2}$
Hence, the correct answer is option: 'A'.