Given #sin60^circ=sqrt3/2# and #cos60^circ=1/2#, how do you find #tan60^circ#?
1 Answer
Aug 10, 2018
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#
