Q.

If the sides of a triangle are in AP, and the greatest angle of the triangle is double the smallest angle, the ratio of the sides of the triangle is

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a

3:4:5

b

4:5:6

c

5:6:7

d

7:8:9

answer is B.

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Detailed Solution

Let the sides of the triangle be a-d, a and a+d, with a > d>0. Clearly, a -d is the smallest side and a + d is the largest side. So, A is the smallest angle and C is the largest angle. It is given that C = 2A. Thus, the angles of the triangle are A, 2A and  π-3A. Applying the law of sines, we obtain  a−dsin⁡A=asin⁡(π−3A)=a+dsin⁡2A⇒a−dsin⁡A=asin⁡3A=a+dsin⁡2A⇒a−dsin⁡A=a3sin⁡A−4sin3⁡A=a+d2sin⁡Acos⁡A⇒a−d1=a3−4sin2⁡A=a+d2cos⁡A⇒3−4sin2⁡A=aa−dand2cos⁡A=a+da−d⇒4cos2⁡A−1=aa−dand2cos⁡A=a+da−d⇒ a+da−d2−1=aa−d⇒a=5dThus, the sides of the triangle are a-d, a, a + d    i.e. 4d, 5d, 6d.Hence, the ratio of the sides of the triangle is 4d :5d: 6d     i.e. 4:5:6.
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