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Question

In a triangle ABC,AD and BE are medians drawn to BC and CA respectively. Given AD=4,DAB=π6 andABE=π3. If the area of triangle ABC is p3q, where p and q are co-prime, then the value of p+q is

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Solution


As, AD=4
So, AG=23AD
AG=23(4)=83

Now, in AGB
BGsin30=AGsin60 (Using sine law)
BG=(sin30sin60)AG
BG=(1/23/2)(83)=833


Now, area (ABC)=3 × area (GAB)
=3(12×BG×AG)
=3×12×833×83
=3233=p3q
So, p=32 and q=3
p+q=32+3=35

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