Question

The point P(21, 28) is on the terminal side of θ. Evaluate sin θ.

Answer 1

Note that for P(21, 28) we get x=21 and y=28.

Consider O(0; 0), then $OP=\sqrt{21^2+28^2}=$ $\sqrt{441+784}=\sqrt{1225}=35$.

sin θ=$\frac{opposite side}{hypotenuse}=\frac{y}{OP}=\frac{28}{35}=\frac{4}{5}.$

Answer: sin θ$=\frac{4}{5}$ .