Year 10A Sine Rule, Cosine Rule and Area of a Triangle Worksheets & Practice

Worked examples

Year 10A sine rule, cosine rule and area: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Using the sine rule to find an unknown side

   In triangle $ABC$, angle $A = 47°$, angle $B = 63°$, and side $a = 14$ cm. Find side $b$, correct to one decimal place.

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   Key idea: When two angles and one opposite side are known, use the sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$.

   1. Write the sine rule with the known values:
      $\frac{14}{\sin 47°} = \frac{b}{\sin 63°}$
   2. Solve for $b$:
      $b = \frac{14 \times \sin 63°}{\sin 47°}$
   3. Evaluate:
      $b = \frac{14 \times 0.8910}{0.7314} \approx \frac{12.474}{0.7314} \approx 17.1 \text{ cm}$

   Answer: $b \approx 17.1$ cm

2. Example 2 · Using the cosine rule to find an unknown angle

   In triangle $PQR$, $p = 9$ cm, $q = 11$ cm, and $r = 7$ cm. Find angle $R$, correct to the nearest degree.

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   Key idea: When all three sides are known, use the cosine rule in the form $\cos R = \dfrac{p^2 + q^2 - r^2}{2pq}$.

   1. Substitute the known values:
      $\cos R = \frac{9^2 + 11^2 - 7^2}{2 \times 9 \times 11}$
   2. Simplify the numerator and denominator:
      $\cos R = \frac{81 + 121 - 49}{198} = \frac{153}{198}$
   3. Divide:
      $\cos R \approx 0.7727$
   4. Find angle $R$:
      $R = \cos^{-1}(0.7727) \approx 39.6° \approx 40°$

   Answer: $R \approx 40°$

3. Example 3 · Area and the cosine rule in a real-life problem

   Two fences meet at a corner post. One fence is 23 m long and the other is 18 m long. The angle between the fences at the corner post is $74°$. A third fence is to be built to join the far ends of the two fences, enclosing a triangular paddock. Find the area of the triangular paddock, correct to the nearest square metre, and the length of the third fence, correct to the nearest metre.

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   Key idea: Use the area formula $\text{Area} = \tfrac{1}{2}ab\sin C$ when two sides and the included angle are known, then use the cosine rule to find the third side.

   1. Label the two known sides $a = 23$ m, $b = 18$ m, and the included angle $C = 74°$.
   2. Calculate the area:
      $\begin{gathered}\text{Area} = \frac{1}{2} \times 23 \times 18 \times \sin 74° \\ = \frac{1}{2} \times 23 \times 18 \times 0.9613 \approx 199.0 \approx 199 \text{ m}^2\end{gathered}$
   3. Apply the cosine rule to find the third side $c$:
      $\begin{gathered}c^2 = 23^2 + 18^2 - 2 \times 23 \times 18 \times \cos 74° \\ = 529 + 324 - 828 \times 0.2756 \\ = 853 - 228.2 = 624.8\end{gathered}$
   4. Take the square root:
      $c = \sqrt{624.8} \approx 25 \text{ m}$

   Answer: Area $\approx 199$ m², third fence $\approx 25$ m

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