Question

in δpqr, r = 630 cm, q = 590 cm and ∠q=164°. find all possible values of ∠r, to the nearest degree.

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that $\frac{\sin R}{r}=\frac{\sin Q}{q}$. We know $r = 630$, $q = 590$, and $\angle Q=164^{\circ}$. Plugging in the values, we get $\frac{\sin R}{630}=\frac{\sin 164^{\circ}}{590}$.

Step2: Solve for $\sin R$

First, calculate $\sin 164^{\circ}$. $\sin 164^{\circ}=\sin(180^{\circ} - 16^{\circ})=\sin 16^{\circ}\approx0.2756$. Then, $\sin R=\frac{630\times\sin 164^{\circ}}{590}=\frac{630\times0.2756}{590}$. Calculate the numerator: $630\times0.2756 = 630\times0.2756 = 173.628$. Then, $\sin R=\frac{173.628}{590}\approx0.2943$.

Step3: Find the possible angle $R$

Now, we find the angle whose sine is approximately $0.2943$. Using the inverse sine function, $R=\sin^{-1}(0.2943)\approx17^{\circ}$. We also need to check if there is another possible angle in the second quadrant (since sine is positive in both first and second quadrants). The other possible angle would be $180^{\circ}-17^{\circ}=163^{\circ}$. But we need to check if this angle is valid. The sum of angles in a triangle is $180^{\circ}$. If $R = 163^{\circ}$ and $Q = 164^{\circ}$, then $R+Q=163^{\circ}+164^{\circ}=327^{\circ}>180^{\circ}$, which is impossible. So the only valid angle is $17^{\circ}$.

Answer:

$\boldsymbol{17^{\circ}}$