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Question

Sides AB and AC in an equilateral triangle ABC with side length 3 is extended to form two rays emanating from the point A as shown in the figure. A point P is chosen outside the triangle ABC and between the two rays such that ABP+BCP=180. If the maximum length of CP is M, then the value of M22 is

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Solution


Given, α+β=180
In BCP,
α60+β+P=180
P=60
Using sine law in BCP
3sin60=xsin(α60)x=23sin(α60)xmax=23MM22=12×4×3=6

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