
ABC is a triangle such that $sin(2A+B) = sin(C-A)=-sin(B+2C)=\dfrac{1}{2}$. If A, B and c are in A.P., then A, B and C are
A. $30^{\circ},60^{\circ},90^{\circ}$
B. $45^{\circ},60^{\circ},75^{\circ}$
C. $45^{\circ},45^{\circ},90^{\circ}$
D. $60^{\circ},60^{\circ},60^{\circ}$
Answer
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Hint: We have a triangle ABC whose angles are in arithmetic progression. We have to find the measures of all the angles of the triangle. To solve this question we mainly use two hints given in the question - one, is the given equation and another, the angles are in A.P. Also, recall the trigonometric results related to sine.
Complete step by step solution: We have a triangle ABC whose angles are in A.P.
Let,
$\angle A=a_{1}$
$\angle B = a_{1}+d$
$\angle C = a_{1}+2d$
We know that the angles of a triangle are supplementary. Therefore, we have
$\angle A + \angle B + \angle C = 180^{\circ}$ ------- (1)
$a_{1} + a_{1} + d + a_{1}+2d = 180^{\circ}$
$3a_{1}+3d = 180^{\circ}$
$3(a_{1}+d) = 180^{\circ}$
$a_{1}+d = \dfrac{180^{\circ}}{3}$
$a_{1}+d = 60^{\circ}$
But we have, $\angle B = a_{1}+d$
$\therefore \angle B = 60^{\circ}$
Now, we have $sin(B+2C) = -\dfrac{1}{2}$
We know that $sin(\pi+x)=-sinx$
$\therefore sin(B+2C) = -sin(30^{\circ}) = sin(180^{\circ}+30^{\circ})$
$sin(B+2C) = sin(210^{\circ})$
$\implies B+2C = 210^{\circ}$
$60^{\circ}+2C = 210^{\circ}$
$2C = 210^{\circ}-60^{\circ}$
$2C = 150^{\circ}$
$C = \dfrac{150^{\circ}}{2}$
$C = 75^{\circ}$
Substituting the values of and in equation (1) we get,
$\angle A+60^{\circ}+75^{\circ} = 180^{\circ}$
$\angle A +135^{\circ} =180^{\circ}$
$\angle A = 180^{\circ}-135^{\circ}$
$\angle A = 45^{\circ}$
Therefore, the measures of the angles A, B and C are $45^{\circ},~60^{\circ}$ and $75^{\circ}$ . So, the answer is Option B.
Note: In this question even though we had three relations for the angles we used only two. We have not used the equation, $sin(2A+B) =\dfrac{1}{2}$ . We can use this equation too. For this we need to keep in mind the equation,$sin(\pi-x)=sinx$ and follow the above steps. Be careful while using the terms of A.P. don’t choose the terms of A.P. as $a_{1},a_{2},a_{3}$ because this is of use with the given information.
Complete step by step solution: We have a triangle ABC whose angles are in A.P.
Let,
$\angle A=a_{1}$
$\angle B = a_{1}+d$
$\angle C = a_{1}+2d$
We know that the angles of a triangle are supplementary. Therefore, we have
$\angle A + \angle B + \angle C = 180^{\circ}$ ------- (1)
$a_{1} + a_{1} + d + a_{1}+2d = 180^{\circ}$
$3a_{1}+3d = 180^{\circ}$
$3(a_{1}+d) = 180^{\circ}$
$a_{1}+d = \dfrac{180^{\circ}}{3}$
$a_{1}+d = 60^{\circ}$
But we have, $\angle B = a_{1}+d$
$\therefore \angle B = 60^{\circ}$
Now, we have $sin(B+2C) = -\dfrac{1}{2}$
We know that $sin(\pi+x)=-sinx$
$\therefore sin(B+2C) = -sin(30^{\circ}) = sin(180^{\circ}+30^{\circ})$
$sin(B+2C) = sin(210^{\circ})$
$\implies B+2C = 210^{\circ}$
$60^{\circ}+2C = 210^{\circ}$
$2C = 210^{\circ}-60^{\circ}$
$2C = 150^{\circ}$
$C = \dfrac{150^{\circ}}{2}$
$C = 75^{\circ}$
Substituting the values of and in equation (1) we get,
$\angle A+60^{\circ}+75^{\circ} = 180^{\circ}$
$\angle A +135^{\circ} =180^{\circ}$
$\angle A = 180^{\circ}-135^{\circ}$
$\angle A = 45^{\circ}$
Therefore, the measures of the angles A, B and C are $45^{\circ},~60^{\circ}$ and $75^{\circ}$ . So, the answer is Option B.
Note: In this question even though we had three relations for the angles we used only two. We have not used the equation, $sin(2A+B) =\dfrac{1}{2}$ . We can use this equation too. For this we need to keep in mind the equation,$sin(\pi-x)=sinx$ and follow the above steps. Be careful while using the terms of A.P. don’t choose the terms of A.P. as $a_{1},a_{2},a_{3}$ because this is of use with the given information.
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