More about trigonometry - (13.4-13.6)

Understand the sine and cosine formulae

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

    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

  • Theory
  • Examples
    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.
    (Give your answers correct to 3 significant figures if necessary.)

  • Solutions

    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

  • Theory
  • Examples
    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.
    (Give your answers correct to 1 decimal place.)
  • Solutions

    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 \(\ 3.7^\circ \).