Answer to Question #90785 in Trigonometry for Unknown287159

Question

Solve triangle ABC given that:A=60° ,AB=12,2cm and BC=14,5.

Answer 1

We can use The Law of Sines first to find angle C:

$$\frac{AB}{\sin{C}}=\frac{BC}{\sin{A}},$$

$$\frac{12.2}{\sin{C}}=\frac{14.5 \cdot 2}{\sqrt{3}},$$ $$C=\arcsin{\frac{61\sqrt{3}}{145}}.$$

The other angle C might be

$$180 \degree - \arcsin{\frac{61\sqrt{3}}{145}} \approx 133.23 \degree,$$

but this is impossible, since the sum of the angles A and C is more than 180°.

Use "the three angles add to 180°" to find angle B:

$$B = 180\degree - A - C = 120\degree - C.$$

Now we can use The Law of Sines again to find AC:

$$\frac{AC}{\sin{B}}=\frac{BC}{\sin{A}},$$

$$\sin{B} = \sin{(120\degree - C)} = \\ \sin{120\degree}\cos{C}-\cos{120\degree}\sin{C} = \\ \frac{\sqrt{3}}{2} \sqrt{1-\left(\frac{61\sqrt{3}}{145}\right)^2} - \left(-\frac{1}{2}\right) \cdot \frac{61\sqrt{3}}{145} = \frac{\sqrt{3}}{290}\left(\sqrt{9862}+61\right),$$

$$AC = \frac{14.5 \cdot 2}{\sqrt{3}} \sin{B} = \frac{1}{10} \left(\sqrt{9862}+61\right).$$

Answer:

$$C=\arcsin{\frac{61\sqrt{3}}{145}} \approx 46.77 \degree,$$ $$B = 120\degree - C \approx 73.23 \degree,$$ $$AC = \frac{1}{10} \left(\sqrt{9862}+61\right) \approx 16.03.$$