Sine Rule

Feedback: Not quite! To apply the Sine Rule, you need to know at least one pair of an angle and its opposite side in the triangle.

Feedback: Not quite! In the Sine Rule formula, $a$, $b$, and $c$ represent the sides of the triangle, and A, B, and C are the angles opposite those sides.

Feedback: Not quite! Using the Sine Rule, $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$. Substituting the known values, $\frac{5}{\sin(30^\circ)} = \frac{b}{\sin(70^\circ)}$. Since $\sin(30^\circ) = 0.5$ and $\sin(70^\circ) \approx 0.94$, we cross-multiply to get $b \approx \frac{5 \times 0.94}{0.5} \approx 9.4$ cm.

Feedback: Not quite! Using the Sine Rule, $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$. Substituting the known values, $\frac{6}{\sin(45^\circ)} = \frac{b}{\sin(60^\circ)}$. Since $\sin(45^\circ) \approx 0.7$ and $\sin(60^\circ) \approx 0.9$, we cross-multiply to get $b \approx \frac{6 \times 0.9}{0.7} \approx 7.7$ cm.

Feedback: Not quite! Using the Sine Rule, $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$. Substituting the known values, $\frac{4}{\sin(30^\circ)} = \frac{b}{\sin(100^\circ)}$. Since $\sin(30^\circ) = 0.5$ and $\sin(100^\circ) \approx 1.0$, we cross-multiply to get $b \approx \frac{4 \times 1.0}{0.5} \approx 8.0$ cm.

Feedback: Not quite! Using the Sine Rule, $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$. Substituting the known values, $\frac{5}{\sin(A)} = \frac{10}{\sin(120^\circ)}$. Since $\sin(120^\circ) \approx 0.866$, we can rearrange the equation to get $\sin(A) \approx \frac{5 \times 0.866}{10} \approx 0.433$. Finally, using the inverse sine function, we find that angle $A \approx 25.8^\circ$.