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−dsinA=asin(π−3A)=a+dsin2A⇒a−dsinA=asin3A=a+dsin2A⇒a−dsinA=a3sinA−4sin3A=a+d2sinAcosA⇒a−d1=a3−4sin2A=a+d2cosA⇒3−4sin2A=aa−dand2cosA=a+da−d⇒4cos2A−1=aa−dand2cosA=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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