Question

The angle 01 is located in Quadrant II, and cos(01)= -12/19 What is the value of sin(01)?

Answer 1

Given that angle θ1 is located in Quadrant II; and

$\cos (\theta_1)=-\frac{12}{19}$

Recall from trigonometric identity:

$\begin{gathered} \cos \theta=\frac{\text{Adjacent}}{\text{Hypotenuse}} \\ \implies\text{Adjacent}=-12 \\ \text{Hypotenuse}=19 \end{gathered}$

We find the value of Opposite using the Pythagoras theorem.

$\begin{gathered} \text{Hyp}^2=\text{Opp}^2+\text{Adj}^2 \\ 19^2=\text{Opp}^2+(-12)^2 \\ \text{Opp}^2=19^2-(-12)^2 \\ \text{Opp}^2=361-144 \\ \text{Opp}^2=217 \\ \text{Opp}=\sqrt[]{217} \end{gathered}$

Therefore, the value of sin(θ1) will be:

$\begin{gathered} \sin (\theta_1)=\frac{\text{Opposite}}{\text{Hypotenuse}} \\ =\frac{\sqrt[]{217}}{19} \end{gathered}$

Note: Sine is positive in Quadrant II.