Mathematics 10 — Exercise 7.2 · Q2 (iv)

Full question

Calculate the area of triangle ABC using the following measurements.

Step-by-step solution

1. Angles of a triangle sum to 180°. Subtract both known angles.

   $$C=180-62.0-69.0$$

2. Calculate the remaining angle.

   $$C=49.0$$

3. Rearrange the sine law to isolate the missing side.

   $$b=\frac{10.0\times \sin \left(\mathrm{69.0°}\right)}{\sin \left(\mathrm{62.0°}\right)}$$

4. Evaluate both sines and multiply the known side by the numerator sine.

   $$b=\frac{9.3358043}{0.88294759}$$

5. Divide by the denominator sine.

   $$b=10.573452$$

6. Rearrange the sine law to isolate the missing side.

   $$c=\frac{10.0\times \sin \left(\mathrm{49.0°}\right)}{\sin \left(\mathrm{62.0°}\right)}$$

7. Evaluate both sines and multiply the known side by the numerator sine.

   $$c=\frac{7.5470958}{0.88294759}$$

8. Divide by the denominator sine.

   $$c=8.5476147$$

9. Use half the product of two sides and the sine of their INCLUDED angle.

   $$\mathrm{area}=\frac{105.73452\times \sin \left(\mathrm{49.0°}\right)}{2}$$

10. Evaluate the sine first.

    $$\sin \left(\mathrm{49.0°}\right)=0.75470958$$

11. Multiply the two side lengths.

    $$10.0\times 10.573452=105.73452$$

12. Multiply by the sine, then divide by 2.

    $$\frac{79.798858}{2}$$

13. The area is in square centimetres.

    $$\mathrm{area\; (cm²)}=39.899429$$

Answer

The final answer is shown below.