Mathematics 10 — Exercise 7.3 · Q4

Full question

Square-based pyramid ABCDE has AE=BE=CE=DE=12 cm and AB=15 cm. Find angle DEB to a whole degree.

Step-by-step solution

1. Opposite corners B and D of a square of side 15 cm have BD²=15²+15²=450.

   $$\mathrm{BD²}=225+225$$

2. In triangle DEB, DE=EB=12 cm. Apply cosine law at E.

   $$\cos \left(\mathrm{DEB°}\right)=\frac{{12}^2+{12}^2-450}{2\times 12\times 12}$$

3. For multiplication of fractions, multiply the numerators together and the denominators together.

   $$\frac{\left(-162\right)\times 1}{1\times 288}$$

4. The numerator and denominator have this greatest common divisor. Divide BOTH by it.

   $$\mathrm{GCD}=18$$

5. Show the two divisions explicitly.

   $$\frac{-\frac{162}{18}}{\frac{288}{18}}$$

6. Evaluate the numerator division.

   $$-\frac{162}{18}=-9$$

7. Evaluate the denominator division.

   $$\frac{288}{18}=16$$

8. Write the reduced fraction.

   $$-\frac{9}{16}$$

9. Calculate this operation before combining the remaining terms.

   $$\left(-162\right)\times \frac{1}{288}=-\frac{9}{16}$$

10. Calculate numerator: 144+144−450=−162. Denominator: 2×12×12=288.

    $$-\frac{162}{288}$$

11. Reduce the ratio and take inverse cosine.

    $$124.22887$$

12. Round to a whole degree.

Answer

The final answer is shown below.