SIHS REVISION EXERCISES FOR END OF S4,S5&S6 Mathematics Topic area: Trigonometry

The formula for angle A is:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

Substitute the given values:

$$\cos A = \frac{19^2 + 14^2 - 8^2}{2 \times 19 \times 14}$$ $$\cos A = \frac{361 + 196 - 64}{532}$$ $$\cos A = \frac{493}{532}$$ $$A = \arccos\left(\frac{493}{532}\right) \approx 22.00^\circ$$

The formula for angle B is:

$$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$

Substitute the given values:

$$\cos B = \frac{8^2 + 14^2 - 19^2}{2 \times 8 \times 14}$$ $$\cos B = \frac{64 + 196 - 361}{224}$$ $$\cos B = \frac{-101}{224}$$ $$B = \arccos\left(\frac{-101}{224}\right) \approx 116.80^\circ$$

The formula for angle C is:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

Substitute the given values:

$$\cos C = \frac{8^2 + 19^2 - 14^2}{2 \times 8 \times 19}$$ $$\cos C = \frac{64 + 361 - 196}{304}$$ $$\cos C = \frac{229}{304}$$ $$C = \arccos\left(\frac{229}{304}\right) \approx 41.19^\circ$$

The three angles of the triangle are:

$$A \approx \boxed{22.00^\circ}$$ $$B \approx \boxed{116.80^\circ}$$ $$C \approx \boxed{41.19^\circ}$$

(Check: $22.00^\circ + 116.80^\circ + 41.19^\circ = 179.99^\circ$, which is approximately $180^\circ$).