More about trigonometry  (13.413.6)
Understand the sine and cosine formulae

Theory

Examplesa) 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.
(Give your answers correct to 1 decimal place if necessary.) 
Solutions\(\begin{array}{1} a)\qquad B = 180^\circ  A  C\\ = 180^\circ  108^\circ  40{^\circ {}}\\ = 32{^\circ {}}\end{array}\)
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

Theory

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

SolutionsLet 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 2dimensional problems

Theory

ExamplesIn 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.
(Give your answers correct to 3 significant figures if necessary.)

Solutionsa) 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 3dimensional problems

Theory

ExamplesIn 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.
(Give your answers correct to 1 decimal place.) 
Solutionsa) 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 \(\ 3.7^\circ \).