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Trigonometry

Question 4 of 9

Sine and Cosine Rules

Question

Solve triangle \(ABC\) if \(a = 15\), \(b = 26\), \(\angle A = 29^\circ\). Find \(c\), \(\angle B\) and \(\angle C\). (Round your answer to the nearest whole number)

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Condition for 2 solutions: \(\angle A\) is an acute angle and \(a \lt b\) where \(a \gt b \sin A\).

\[ (15 \gt 26 \sin 29 = 15 \gt 13) \]

The sketch of the triangle shows that side \(a\) may be drawn in two possible positions to complete the triangle.

Sketch of the two possible triangles, drawn to scale. Vertex A is at the bottom left, with the 29° angle marked between side b and the base direction; side b = 26 rises from A to C. From C, side a = 15 can be drawn down to the base in two positions: at B₂, nearer to A, and at B₁, further from A. The base is drawn dashed because its length is not fixed by the given data; the two descending segments, each labelled a = 15, show the two positions in which side a completes the triangle.

Hence the two triangles satisfy the given conditions: triangle \(AB_1C_1\) and triangle \(AB_2C_2\).

  1. Solve triangle \(AB_1C_1\): use the sine rule to solve for \(\angle B_1\)
    Triangle AB₁C₁, drawn to scale: the first of the two triangles. The 29° angle is marked at A; side b = 26 runs from A up to C₁; side a = 15 runs from C₁ down to B₁; the base from A to B₁ is side c₁, which is to be found.
    \[ \frac{a}{\sin A} = \frac{b}{\sin B_1} \]

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

    \[ \frac{15}{\sin 29} = \frac{26}{\sin B_1} \]
  2. Cross multiply and solve for \(\sin B_1\)
    \[ \sin B_1 = \frac{26 \sin 29}{15} \]
    \[ \sin B_1 = 0.8403 \]
  3. Solve for angle \(B_1\)
    \[ \angle B_1 = 57.18^\circ \approx 57^\circ \]
  4. Solve for angle \(C_1\)
    \[ \angle C_1 = 180 - (29 + 57) = 94^\circ \]
  1. Use the sine rule to solve for \(c_1\)
    \[ \frac{a}{\sin A} = \frac{c}{\sin C_1} \]

    Substitute the values of \(a\), \(\angle A\) and \(\angle C_1\):

    \[ \frac{15}{\sin 29} = \frac{c}{\sin 94} \]
  2. Cross multiply and solve for \(c_1\)
    \[ c_1 = \frac{15 \sin 94}{\sin 29} \]
    \[ c_1 = 30.86 \approx 31 \]
  3. Solve triangle \(AB_2C_2\): solve for angle \(B_2\) (straight angles \(= 180^\circ\))
    Triangle AB₂C₂, drawn to scale: the second of the two triangles. The 29° angle is marked at A; side b = 26 runs from A up to C₂; side a = 15 runs from C₂ down to B₂; the base from A to B₂ is side c₂, which is to be found. A dashed ray extends the base beyond B₂, and the 57° exterior angle between that ray and side a is marked; the interior angle of the triangle at B₂ is its supplement.
    \[ \angle B_2 = 180^\circ - 57^\circ = 123^\circ \]
  4. Solve for angle \(C_2\) (sum of interior angles of a triangle \(= 180^\circ\))
    \[ \angle C_2 = 180^\circ - (29^\circ + 123^\circ) = 28^\circ \]
  5. Use the sine rule to solve for \(c_2\)
    \[ \frac{a}{\sin A} = \frac{c}{\sin C} \]

    Substitute the values of \(a\), \(\angle A\) and \(\angle C_2\):

    \[ \frac{15}{\sin 29} = \frac{c}{\sin 28} \]
  6. Cross multiply and solve for \(c_2\)
    \[ c_2 = \frac{15 \sin 28}{\sin 29} \]
    \[ c_2 = 14.53 \approx 15 \]
  7. Final answer

    Therefore, \(c_1 = 31\), \(\angle B_1 = 57^\circ\), \(\angle C_1 = 94^\circ\) and \(c_2 = 15\), \(\angle B_2 = 123^\circ\), \(\angle C_2 = 28^\circ\). That is, in the first triangle, side c₁ is 31, angle B₁ is 57 degrees and angle C₁ is 94 degrees; in the second triangle, side c₂ is 15, angle B₂ is 123 degrees and angle C₂ is 28 degrees.