Question

Given `sin60^circ=sqrt3/2` and `cos60^circ=1/2`, how do you find `tan60^circ`?

Answer 1

`tan60^@=sqrt3`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)tantheta=sintheta/costheta`

`"thus"`

`tan60^@=sin60^@/cos60^@`

`color(white)(xxxxx)=(sqrt3/2)/(1/2)=sqrt3/cancel(2)xxcancel(2)/1=sqrt3`