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In the triangle ABC, AB = 16 cm, AC = 13 cm, angle ABC = 50° and angle BCA = x° Find the two possible values for x, giving your answers to one decimal place. - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 3

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In the triangle ABC, AB = 16 cm, AC = 13 cm, angle ABC = 50° and angle BCA = x° Find the two possible values for x, giving your answers to one decimal place.

Worked Solution & Example Answer:In the triangle ABC, AB = 16 cm, AC = 13 cm, angle ABC = 50° and angle BCA = x° Find the two possible values for x, giving your answers to one decimal place. - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 3

Step 1

Find the value of sin(x)

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Answer

Using the sine rule: asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} Here, we know:

  • a = AC = 13 cm
  • b = AB = 16 cm
  • angle ABC = 50°

Substituting the values: 16sin⁡50°=13sin⁡x\frac{16}{\sin 50°} = \frac{13}{\sin x}

Rearranging gives: sin⁡x=13⋅sin⁡50°16\sin x = \frac{13 \cdot \sin 50°}{16}

Calculating the value: sin⁡x=13⋅0.766016≈0.943\sin x = \frac{13 \cdot 0.7660}{16} \approx 0.943

Step 2

Find the possible values for x

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Answer

Now, solving for x gives:

  1. First possible solution:

    • x=sin⁡−1(0.943)≈70.5°x = \sin^{-1}(0.943) \approx 70.5°

  2. Second possible solution (using the property of sine where it's symmetrical):

    • x=180°−70.5°=109.5°x = 180° - 70.5° = 109.5°

Therefore, the two possible values of x are approximately 70.5° and 109.5°.

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