If 5 tan θ-12=0, then the value of sin θ is
(2)
Given, 5tanθ-12=0⇒tanθ=125=pbLet p=12k, b=5kh=p2+b2=(12k)2+(5k)2=144k2+25k2=169k2=13ksinθ=ph=12k13k=1213