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T3 The sine rule

 

The Sine rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given:

  1. One side and any two angles or

  2. Two sides and an angle opposite one of the given sides.

The Sine Rule

In the triangle \(ABC\) below:

  • angles \(A,B,C\), are the angles at the vertices \(A,B,C\) respectively

  • \(a,b,c\) are the side lengths opposite the angles \(A,B,C\) respectively.

a triangle with sides a, b and c and opposite angles capital A, capital B and capital C respectively

The sine rule states:

\[\begin{align*} \frac{a}{\sin A} & =\frac{b}{\sin B}=\frac{c}{\sin C} \end{align*}\]

or

\[\begin{align*} \frac{\sin A}{a} & =\frac{\sin B}{b}=\frac{\sin C}{c} \end{align*}\]

Examples

  1. In triangle \(PQR\) find:

a triangle with sides p, q and r where r=15 and opposite angles capital P, capital Q and capital R where angle P=70 and angle R=30

  1. the side length \(p\)
  2. the side length \(q\).
    Solution a.
    Use the sine rule in the form \[\begin{align*} \frac{p}{\sin P} & =\frac{r}{\sin R}\\ \frac{p}{\sin70^{\circ}} & =\frac{15}{\sin30^{\circ}}\\ p & =\frac{15\times\sin70^{\circ}}{\sin30^{\circ}}\\ p & =28.2\;. \end{align*}\] The side length is \(p=28.2\,cm.\)
    Solution b)
    Angle \(Q\) is found using the fact that the sum of the three interior angles of a triangle add to \(180^{\circ}\): \[\begin{align*} \angle Q & =180^{\circ}-\left(70^{\circ}+30^{\circ}\right)\\ & =80^{\circ}. \end{align*}\] From the Sine rule \[\begin{align*} \frac{q}{\sin Q} & =\frac{r}{\sin R}\\ \frac{q}{\sin80^{\circ}} & =\frac{15}{\sin30^{\circ}}\\ q & =\frac{15\times\sin80^{\circ}}{\sin30^{\circ}}\\ q & =29.6\;. \end{align*}\] The length of the side \(q=29.6\,cm.\)
  1. In triangle \(ABC\) find:

a triangle with sides a, b and c, where b=20 and c=12 and angles capital A, capital B anc capital C, angle capital B is 126

  1. Angle C
  2. Angle A
  3. Side length a.
    Solution a.
    Use the sine rule in the form \[\begin{align*} \frac{\sin A}{a} & =\frac{\sin B}{b}=\frac{\sin C}{c} \end{align*}\] The relevant part of the formula is \[\begin{align*} \frac{\sin B}{b} & =\frac{\sin C}{c}\\ \frac{\sin126^{\circ}}{20} & =\frac{\sin C}{12}\\ \frac{\sin126^{\circ}}{20}\times12 & =\sin C\\ \sin C & =0.485\\ C & =\sin^{-1}0.485\\ C & =29^{\circ}. \end{align*}\] Solution b.
    \[\begin{align*} \angle A & =180^{\circ}-\left(126^{\circ}+29^{\circ}\right)\\ & =25^{\circ}. \end{align*}\] Solution c.
    Use the sine rule in the form \[\begin{align*} \frac{a}{\sin A} & =\frac{b}{\sin B}=\frac{c}{\sin C} \end{align*}\] The relevant part of the formula is \[\begin{align*} \frac{a}{\sin A} & =\frac{b}{\sin B}\\ \frac{a}{\sin25^{\circ}} & =\frac{20}{\sin126^{\circ}}\\ a & =\frac{20}{\sin126^{\circ}}\times\sin25^{\circ}\\ a & =10.4m \end{align*}\]
  1. Given a a triangle \(ABC\) with angle \(A=30^{\circ}\), side \(c=12\) cm and side \(a=8\) cm, find angle \(C\).
    Solution:
    In this case there are two possible solutions as shown below.

two triangles with Labelled A, B, C. In both cases the angle A is 30, side c is 30, side a is 8, but in one case the angle B is acute, in the other it is obtuse

This is called the ambiguous case of the sine rule.
To solve the problem, use the sine rule in the form: \[\begin{align*} \frac{\sin A}{a} & =\frac{\sin C}{c}\\ \frac{\sin30^{\circ}}{8} & =\frac{\sin C}{12}\\ \frac{\sin30^{\circ}}{8}\times12 & =\sin C\\ \sin C & =\frac{1}{16}\times12\\ & =0.75\\ C & =\sin^{-1}\left(0.75\right)\\ & =48.6^{\circ}\textrm{or }131.4^{\circ}. \end{align*}\] If \(C=48.6^{\circ},\) this solution gives the triangle \(ABC_{2}.\)
If \(C=131.4^{\circ}=\left(180^{\circ}-48.6^{\circ}\right)\), this solution gives the triangle \(ABC_{1}.\)


Note: In any non-right-angled triangle, where two sides and the non-included angle are given, check for the ambiguous case.

If the angle is acute and the length of the side adjacent to the angle is greater than the length of the side opposite the angle there may be 2 possible solutions.


Exercise 1

For the following triangles find the unknown sides.

a triangle with sides a, b and c with a=10 and angles capital A, capital B and capital C with angle A=30 and angle B=65

a triangle with sides p, q and r opposite to angles, capital P, Capital Q and capital R respectively. Side p is 38. Angle P is 45 and angle Q is 85

a triangle with sides r, s and t opposite to angles, capital R, capital S and capital T respectively. Side s is 27, angle S is 130, angle T is 15

  1. \(b=18.1\), \(c=19.9\)

  2. \(q=53.5\), \(r=41.2\)

  3. \(t=9.1\), \(r=20.2\)

Exercise 2

For the following triangles find all unknown angles and sides.

a triangle with sides l, m and n opposite to angles, capital L, capital M and capital N respectively. Side l is 5 and the side m is 8.5, angle M is 80

a triangle with sides a, b and c opposite to angles, capital A, capital B and capital C respectively. Side b is 4.7, side c is 6.4, angle C is 65

  1. Given triangle \(ABC\), where angle \(A=35,\) side \(a=16\) and side \(c=21,\) find the magnitude of angles \(B\) and \(C\) and the length of side \(b\).
  1. \(L=35.4^{\circ}\) , \(N=64.6^{\circ}\), \(n=7.8\)

  2. \(B=41.7^{\circ}\), \(A=73.3^{\circ}\), \(a=6.8\)

  3. Ambiguous case: \(C=131.2^{\circ}\), \(B=13.8^{\circ}\), \(b=6.7\) or \(C=48.8\), \(B=96.2\), \(b=27.7\)

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