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Trigonometry

Question 1 of 9

Sine and Cosine Rules

Question

Given triangle \(ABC\), where \(b = 3\), \(\angle A = 35^\circ\), \(\angle C = 85^\circ\). Find \(a\), \(c\), \(\angle B\)

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 A is marked 35° and the angle at C is marked 85°; an unmarked arc at B indicates angle B, which is to be found. Side AC (side b, opposite B) is labelled 3; side BC (side a, opposite the 35° angle at A) and the base AB (side c, opposite the 85° angle at C) are the sides to be found, labelled a 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 \(180^\circ\):

    \[ \angle B = 180^\circ - (35^\circ + 85^\circ) = 60^\circ \]
  2. Using the sine rule, substitute the values of \(b\), \(\angle A\) and \(\angle B\)
    \[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
    \[ \frac{a}{\sin 35^\circ} = \frac{3}{\sin 60^\circ} \]
  3. Cross multiply and solve for \(a\)
    \[ a = \frac{3 \sin 35^\circ}{\sin 60^\circ} \]
    \[ a = 1.99 \approx 2 \]
  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 \(\angle C\):

    \[ \frac{1.99}{\sin 35^\circ} = \frac{c}{\sin 85^\circ} \]
  5. Cross multiply and solve for \(c\)
    \[ c = \frac{1.99 \sin 85^\circ}{\sin 35^\circ} \]
    \[ c = 3.46 \approx 3 \]
  6. Final answer

    Therefore, \(a = 2\), \(c = 3\) and \(\angle B = 60^\circ\). That is, side a is 2, side c is 3, and angle B is 60 degrees.