Find the cosine of the angle at the vertex of an isoceles triangle having the greatest area for the given constant length e of the median drawn to its lateral side.

Options

A. 0.4
B. 0.5
C. 0.6
D. 0.8

Answer

4

Explanation

angle ABC is isosceles with AB=AC
BE and CF are medians
Area of triangle
`triangle=1/2 bc sin theta`
`=1/2 c^(2) sin theta (as b=c)`
In `triangle ABE` using consine rule
`l^(2)=c^(2)+(b^(2))/(4)-bc cos theta`
`=(5c^(2))/(4)-c^(2)cos theta`
`therefore c^(2)=(4l^(2))/(5-4cos theta)`
`therefore triangle=2l^(2)(sin theta)/(5-4cso theta)`
`therefore (d triangle)/(d theta)=(2^(2))(5-4cos theta)cos theta - sin theta (4 sin theta)/(5-4 sin theta)^(2)`
`=(2^(2))(5cos theta -4)/(5-4 sin theta)^(2)`
`therefore For triangle` to be maximum `cos theta =0.8`