Question

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.
Figure

Answer 1

From the given right-angled triangle, we have:
$$\begin{gathered} \frac{BC}{AC}=\sin {30}^{\mathrm{o}} \\ \Rightarrow \frac{BC}{20}=\frac{1}{2} \\ \Rightarrow BC=\frac{20}{2}=10\mathrm{cm} \\ \mathrm{Also},\frac{AB}{AC}=\cos {30}^{\mathrm{o}} \\ \Rightarrow \frac{AB}{20}=\frac{\sqrt{3}}{2} \\ \Rightarrow AB=\left(20\times \frac{\sqrt{3}}{2}\right)=10\sqrt{3}\mathrm{cm} \\ \therefore \mathrm{BC}=10\mathrm{cm}\mathrm{and}\mathrm{AB}=10\sqrt{3}\mathrm{cm} \end{gathered}$$