Question

In a ABC triangle the medians AD and BE are drawn such that AD =4 angle BAD is 30° and angle ABE is 60° then find the area of the triangle

Answer 1

consider the point of intersection of AD and BE as O.now as BAD=30 and ABE is 60therefore AOB=90.also we know that median divides a triangle into two equal areas.area of ABD=area of ADC also O is centroid.therefore AO=(2*4/3)=8/3 sin(60)=AO/ABAB=16/(3*root(3)) now you know AD,AB  and angle between the two i.e. 30 so area(ABD) =1/2*AB*AD*sin(theta)= 16/(3*root(3))(theta=30)therefore, the total area =2(area(ABD))=32/(3*root(3))

Answer 2

Answer:

The answer to this question is 32/(3*(3^(1÷2)))