Question

In the adjoining figure, $\angle \mathrm{B}=90°,\angle \mathrm{BAC}=\theta °,\mathrm{BC}=\mathrm{CD}=4\mathrm{cm}\mathrm{and}\mathrm{AD}=10\mathrm{cm}.\mathrm{Find}\left(\mathrm{i}\right)\sin \theta \mathrm{and}\left(\mathrm{ii}\right)\cos \theta .$

Answer 1

In $∆$ABD,

Using Pythagoras theorem, we get

$$\begin{gathered} \mathrm{AB}=\sqrt{{\mathrm{AD}}^2-{\mathrm{BD}}^2} \\ =\sqrt{{10}^2-8^2} \\ =\sqrt{100-64} \\ =\sqrt{36} \\ =6\mathrm{cm} \end{gathered}$$

Again,

In $∆$ABC,

Using Pythagoras therem, we get

$$\begin{gathered} \mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2} \\ =\sqrt{6^2+4^2} \\ =\sqrt{36+16} \\ =\sqrt{52} \\ =2\sqrt{13}\mathrm{cm} \end{gathered}$$

Now,

$$\begin{gathered} \left(\mathrm{i}\right)\sin \theta =\frac{\mathrm{BC}}{\mathrm{AC}} \\ =\frac{4}{2\sqrt{13}} \\ =\frac{2}{\sqrt{13}} \\ =\frac{2\sqrt{13}}{13} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\cos \theta =\frac{\mathrm{AB}}{\mathrm{AC}} \\ =\frac{6}{2\sqrt{13}} \\ =\frac{3}{\sqrt{13}} \\ =\frac{3\sqrt{13}}{13} \end{gathered}$$