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Trigonometry

Question 2 of 9

Sine and Cosine Rules

Question

Given triangle \(ABC\), where \(a = 70.5\), \(\angle B = 62^\circ\), \(\angle C = 55^\circ\). Find \(b\), \(c\), \(\angle A\)

Triangle ABC drawn to scale from the given data. Vertex A is at the bottom left, vertex B at the bottom right and vertex C at the top. The angle at B is marked 62° and the angle at C is marked 55°; an unmarked arc at A indicates angle A, which is to be found. Side CB (side a, opposite A) is labelled 70.5; side AC (side b, opposite the 62° angle at B) and the base AB (side c, opposite the 55° angle at C) are the sides to be found, labelled b and c.
Show solution Hide solution Fully worked — 6 steps

This is an example of Case 1: ASA or AAS.

  1. Solve for the third angle

    The sum of the angles of a triangle is \(180^\circ\):

    \[ \angle A = 180^\circ - (62^\circ + 55^\circ) = 63^\circ \]
  2. Use the sine rule to solve for \(b\)
    \[ \frac{a}{\sin A} = \frac{b}{\sin B} \]

    Substitute the values of \(a\), \(\angle A\) and \(\angle B\):

    \[ \frac{70.5}{\sin 63} = \frac{b}{\sin 62} \]
  3. Cross multiply and solve for \(b\)
    \[ b = \frac{70.5 \sin 62}{\sin 63} \]
    \[ b = 69.86 \approx 69.9 \]
  4. Use the sine rule to solve for \(c\)
    \[ \frac{a}{\sin A} = \frac{c}{\sin C} \]

    Substitute the values of \(a\), \(\angle A\) and \(\angle C\):

    \[ \frac{70.5}{\sin 63} = \frac{c}{\sin 55} \]
  5. Cross multiply and solve for \(c\)
    \[ c = \frac{70.5 \sin 55}{\sin 63} \]
    \[ c = 64.81 \approx 64.8 \]
  6. Final answer

    Therefore, \(b = 69.9\), \(c = 64.8\) and \(\angle A = 63^\circ\). That is, side b is 69.9, side c is 64.8, and angle A is 63 degrees.