If one side of a triangle is double the other, and the angles on opposite sides differ by `60^0,` then the triangle is equilateral (b) obtuse angled (c) right angled (d) acute angled

Options

A. obtuse abgled
B. acute angled
C. isosceles
D. right angled

Answer

D

Explanation

Given, `A - B = 60^(@)`

By sine rule `(2a)/(sinA) = (a)/(sinB)`
`rArr " "sin A - 2 sin B = 0`
`rArr " "sin (60^(@) + B) - 2 sin B = 0`
`rArr sin (sqrt(3))/(2) cos B + (1)/(2) sin B -2 sin B = 0`
`rArr (sqrt(3))/(2) cos B -(3)/(2) sin B =0`
`rArrsqrt(3) ((1)/(2) cos B - (sqrt(3))/(2)sinB)=0`
`rArr sqrt(3)[cos(60^(@) + B)] = 0`
`rArr " "60^(@) + B = 90^(@)`
`rArr" " B = 30^(@)`
`rArr" "A = 90^(@)`
Hence, it is right angled triangle .