Mathematics 10 — Exercise Review · Q2 (i)

Full question

Calculate the area of triangle ABC using the following measurements.

Step-by-step solution

1. Use cosine law to find this angle’s cosine.

   $$\cos \left(\mathrm{A°}\right)=\frac{{6.0}^2+{8.0}^2-16.0}{2\times 6.0\times 8.0}$$

2. Calculate the squared terms, subtract, then divide by the denominator.

   $$\cos \left(\mathrm{A°}\right)=0.875$$

3. Take inverse cosine in DEGREE mode.

   $$A=28.955024$$

4. Use cosine law to find this angle’s cosine.

   $$\cos \left(\mathrm{B°}\right)=\frac{{4.0}^2+{8.0}^2-36.0}{2\times 4.0\times 8.0}$$

5. Calculate the squared terms, subtract, then divide by the denominator.

   $$\cos \left(\mathrm{B°}\right)=0.6875$$

6. Take inverse cosine in DEGREE mode.

   $$B=46.567463$$

7. Subtract A and B from 180° to find C.

   $$C=104.47751$$

8. For three given sides use Heron’s formula. First find the semiperimeter.

   $$s=\frac{4.0+6.0+8.0}{2}$$

9. Add the three sides, then divide by 2.

   $$s=9.0$$

10. Subtract this side from s to obtain one Heron factor.

    $$9.0-4.0=5.0$$

11. Subtract this side from s to obtain one Heron factor.

    $$9.0-6.0=3.0$$

12. Subtract this side from s to obtain one Heron factor.

    $$9.0-8.0=1.0$$

13. Multiply all four factors, then take their positive square root.

    $$\mathrm{area}=\sqrt{9.0\times 5.0\times 3.0\times 1.0}$$

14. The area is in square centimetres.

    $$\mathrm{area\; (cm²)}=11.61895$$

Answer

The final answer is shown below.