Question

1. solve for the length of side ad and angle c in the given figure. give your answers to the nearest tenth of a unit.

(figure with right angles at a, d, and a dashed line from b to the vertical side at c, with ab = 2 cm, ad (labeled as 5.7cm? maybe a note), bc = 5 cm, and the vertical side from d to c is 6 cm)

Explanation:

Step1: Identify right triangle for AD

Let $AD = x$, draw $BE \perp CD$ (E on CD), so $EC = 6 - 2 = 4$ cm, $BE = AD = x$.

Step2: Use Pythagorean theorem

In $\triangle BEC$, $BE^2 + EC^2 = BC^2$
$$x^2 + 4^2 = 5^2$$
$$x^2 = 25 - 16 = 9$$
$$x = 3$$
Wait, correction: $EC = CD - BE_{vertical}$, actually $AE=2$, $CD=6$, so $EC=6-2=4$, $BC=5$. Then $AD=BE=\sqrt{5^2 - 4^2}=\sqrt{9}=3.0$ cm.

Step3: Calculate angle C

Use $\sin C = \frac{BE}{BC}$
$$\sin C = \frac{3}{5} = 0.6$$
$$\angle C = \arcsin(0.6) \approx 36.9^\circ$$

Answer:

Length of $AD = 3.0$ cm, $\angle C \approx 36.9^\circ$