### More about trigonometry - (13.4-13.6)

#### Understand the sine and cosine formulae

• a) In △ABC, $$A = 108^\circ$$, $$C = 40^\circ$$ and $$a = 17$$. Solve △ABC.

b)In △ABC, a=13, b=16 and c=18. Find the largest angle of △ABC.
• $$\begin{array}{1} a)\qquad B = 180^\circ - A - C\\ = 180^\circ - 108^\circ - 40{^\circ {}}\\ = 32{^\circ {}}\end{array}$$
By the sine formula,
$$\begin{array}{1}\frac{b}{{\sin 32^\circ }} = \frac{{17}}{{\sin 108^\circ }}\\b = {\frac{{17\sin 32^\circ }}{{\sin 108^\circ }}^{^{}}}\end{array}$$
$$b= 9.5$$ (corr. to 1 d.p.)

$$\begin{array}{1}\frac{c}{{\sin 40^\circ }} = \frac{{17}}{{\sin 108^\circ }}\\c = {\frac{{17\sin 40^\circ }}{{\sin 108^\circ }}^{^{}}}\end{array}$$
$$c= 11.5$$ (corr. to 1 d.p.)
$$\therefore\ \underline{\underline {B = 32^\circ ,{\rm{ }}b = 9.5{\rm{ }},{\rm{ }}c = 11.5}}$$

b) C must be the largest angle since c is the longest side.
By the cosine formula,
$$\cos C = \frac{{{{13}^2} + {{16}^2} - {{18}^2}}}{{2 \times 13 \times 16}}$$
$$C = 75.9^\circ$$ (corr. to 1 d.p.)
$$\therefore$$ The largest angle is $$\ 75.9\circ$$

#### Understand Heron’s formula

• In the figure, find the area of △ABC. (Give your answer correct to 1 decimal place.)

• Let s cm be half of the perimeter of△ABC.
$$\begin{array}{1}s = \frac{{3 + 6 + 7}}{2}\\ = 8\end{array}$$
Area of△ABC $$= \sqrt {8(8 - 3)(8 - 6)(8 - 7)} {\rm{ c}}{{\rm{m}}^2}$$
$$= \underline{\underline {8.9{\rm{ c}}{{\rm{m}}^2}}}$$ (corr. to 1 d.p.)

#### Use the above formulae to solve 2-dimensional problems

• In the figure, ship B is due south of ship A. The bearings of ship C from ship A and ship B are $$\ S52^\circ$$ Eand $$\ N65^\circ$$E respectively. It is given that $$BC = 800{\rm{ m}}$$, A, B and C lie on the same horizontal plane.

(a) Find ∠ACB.
(b)Find the distance between ship A and ship B.

• a) In△ABC,
$$\begin{array}{1}\angle ACB = 180^\circ - 52^\circ - 65^\circ \\ = {\underline{\underline {63^\circ }} ^{}}\end{array}$$

b) By the sine formula,
$$\begin{array}{1}\frac{{AB}}{{\sin 63^\circ }} = \frac{{800{\rm{ m}}}}{{\sin 52^\circ }}\\AB = \frac{{800\sin 63^\circ }}{{\sin 52^\circ }}{\rm{ }}{{\rm{m}}^{^{^{^{}}}}}\end{array}$$
$$= 905{\rm{ m}}$$ (corr. to 3 sig. fig.)
$$\therefore$$ The distance between ship A and ship B is 905 m.

#### Use the above formulae to solve 3-dimensional problems

• In the figure, ABCD is a horizontal plane. ABFE is a hillside, where the greatest inclination of
it is $$\ 12^\circ$$. Path BE on the hill makes an angle of $$\ 72^\circ$$ with the line of greatest slope BF.
It is given that F is 40 m above the horizontal plane, ABCD, CDEF and ABFE are rectangles.
(a) Find the length of path BE.
(b) Find the inclination of path BE.
• a) In△BCF,
$$\begin{array}{1}BF = \frac{{CF}}{{\sin 12^\circ }}\\ = \frac{{40}}{{\sin 12^\circ }}{\rm{ }}{{\rm{m}}^{^{^{^{}}}}}\end{array}$$
In△BEF,
$$\begin{array}{1}BE = \frac{{BF}}{{\cos 72^\circ }}\\ = \frac{{40}}{{\sin 12^\circ \cos 72^\circ }}{\rm{ }}{{\rm{m}}^{^{^{^{}}}}}\end{array}$$
$$= 622.6{\rm{ m}}$$ (corr. to 1 d.p.)
$$\therefore$$ The length of path BE is 622.6 m.

b) Let $$\theta$$ be the inclination of path BE.
In△BDE,
$$\begin{array}{1}\sin \theta = \frac{{DE}}{{BE}}\\ = \frac{{40}}{{{\textstyle{{40} \over {\sin 12^\circ \cos 72^\circ }}}}}\end{array}$$
$$\theta = 3.7^\circ$$ (corr. to 1 d.p.)
$$\therefore$$ The inclination of path BE is $$\ 3.7^\circ$$.