Question

Given a ∆ABC in which
I. ∠A, ∠B and ∠C are in the ratio 3 : 2 : 1.
II. AB, AC and BC are in the ratio $3:\sqrt{3}:2\sqrt{3}$ and $AB=3\sqrt{3}\mathrm{cm}$.
Is ∆ABC a right triangle?
The question give above has two Statements I and II. Answer the question by using instructions given below:
(a) If the question can be answered by one of the given statements only and not by the other.
(b) If the question can be answered by using either statement along.
(c) If the question can be answered by using both the statements but cannot be answered by using either statement.
(d) If the question cannot be answered even by using both the statements together.

Answer 1

(b)
Let $\angle$A, $\angle$ B and $\angle$C be 3x, 2x and x, respectively.
$$\begin{gathered} \therefore 3x+2x+x=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 6x=180° \\ \Rightarrow x=30° \end{gathered}$$
$$\begin{gathered} \mathrm{Now},\mathrm{we}\mathrm{have}: \\ \angle A=3x=3\times 30°=90° \\ \angle B=2x=2\times 30°=60° \\ \angle C=x=30° \end{gathered}$$

Hence, it is a right triangle at A.
Let the sides of the triangle be $3x\mathrm{cm},\sqrt{3}x\mathrm{cm}\mathrm{and}2\sqrt{3}x\mathrm{cm}.$
Also,
$$\begin{gathered} AB=3\sqrt{3}\mathrm{cm} \\ \mathrm{or},3\mathrm{x}=3\sqrt{3} \\ \mathrm{or},\mathrm{x}=3 \end{gathered}$$
Now,
$$\begin{gathered} AC=\sqrt{3}x=\sqrt{3}\times \sqrt{3}=3\mathrm{cm} \\ BC=2\sqrt{3}\mathrm{x}=2\sqrt{3}\times \sqrt{3}=6\mathrm{cm} \end{gathered}$$

$$\begin{gathered} B{C}^2=6^2=36\mathrm{cm} \\ A{B}^2+A{C}^2={\left(3\sqrt{3}\right)}^2+{\left(3\right)}^2=27+9=36\mathrm{cm} \\ \mathrm{Here}, \\ B{C}^2=A{B}^2+A{C}^2 \\ \mathrm{Thus},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{right}-\mathrm{angled}\mathrm{triangle}. \end{gathered}$$

So, it can be answered using either statement.