Question

In the adjoining figure, $\Delta ABC$ is a right – angled at B and $\angle A=30{}^\circ $. If BC = 6cm, find (i) AB, (ii) AC.

Answer 1

Hint: We will be using the concept of trigonometric ratios to solve the problem. We will be using the trigonometric ratio $\tan \theta $ about A to find AB and $\sin \theta $ about A to find the value of AC.

Complete step-by-step answer:
Now, we know that the, $\Delta ABC$ is a right – angle triangle. So, we will apply trigonometric ratio $\tan \theta $ about A to find the value of AB. So, we have,

$\tan 30{}^\circ =\dfrac{P}{B}$

Where P is perpendicular and B is base. So, we have,

$\tan 30{}^\circ =\dfrac{BC}{AB}$

Now, we know that the value of BC = 6cm and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$. So, we have,

$\begin{align}

  & \dfrac{1}{\sqrt{3}}=\dfrac{6}{AB} \\

 & AB=6\sqrt{3}cm \\

\end{align}$

Now, we will apply $\sin \theta $ about A. So, we have $\sin 30{}^\circ =\dfrac{P}{H}$.

Where P is perpendicular and H is hypotenuse. So, we have,

$\sin 30{}^\circ =\dfrac{BC}{AC}$

Now, we know that the value of $\sin 30{}^\circ =\dfrac{1}{2}$ and BC = 6cm. So, we have,

\[\begin{align}

  & \dfrac{1}{2}=\dfrac{6}{AC} \\

 & AC=12cm \\

\end{align}\]

Hence, the value of (i) \[AB=6\sqrt{3}cm\] (ii) AC = 12cm.



Note: To solve these type of question it is important to note that we have used the trigonometric ratios like,

$\sin \theta =\dfrac{P}{H}$

$\tan \theta =\dfrac{P}{B}$

to solve the problem.