Question

In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°. If BC = 6 cm, find (i) AB, (ii) AC.

Answer 1

From the given right-angled triangle, we have:

$$\begin{gathered} \frac{BC}{AB}=\tan {30}^{\mathrm{o}} \\ \Rightarrow \frac{6}{AB}=\frac{1}{\sqrt{3}} \\ \Rightarrow AB=6\sqrt{3}\mathrm{cm} \\ \mathrm{Also},\frac{BC}{AC}=\sin {30}^o \\ \Rightarrow \frac{6}{AC}=\frac{1}{2} \\ \Rightarrow AC=\left(2\times 6\right)=12\mathrm{cm} \\ \therefore AB=6\sqrt{3}\mathrm{cm}\mathrm{and}AC=12\mathrm{cm} \end{gathered}$$