Question

If the angles of a triangle are ${{30}^{\circ }}$, ${{60}^{\circ }}$ and ${{90}^{\circ }}$, then what is the ratio of the sides opposite to these angles?
(a) $\sqrt{3}:\sqrt{2}:1$
(b) $1:\sqrt{2}:2$
(c) $2:\sqrt{3}:1$
(d) $3:2:1$

Answer 1

Hint: First of all, we will find a relation between any two sides by using sin or cosine function as the triangle is a right-angled triangle. Then using that relation, we will find the value of the third side by using Pythagoras theorem i.e. $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$. In this way we will find values of all the three sides and then using them we will find the ratio of three sides of a triangle.

Complete step-by-step solution:
Let us suppose a triangle ABC having angles ${{30}^{\circ }}$, ${{60}^{\circ }}$ and ${{90}^{\circ }}$, so first of all we will draw a diagram for our simplicity.

Now, from the figure we can see that the $\angle B={{90}^{\circ }}$ and angle made by point C is ${{60}^{\circ }}$. So, from this, we can say that the side AC is hypotenuse to the sides AB and BC.
Now, we know that $\sin \theta =\dfrac{\text{Opposite}}{\text{Hypotenuse}}$, using this rule for point C we will get,
$\sin 60=\dfrac{\text{Opposite}}{\text{Hypotenuse}}$
Where, opposite side to angle made by point C will be AB and the hypotenuse is AC, on substituting these in equation we will get,
$\sin 60=\dfrac{AB}{AC}$
Now, the value of $\sin {{60}^{\circ }}$ is $\dfrac{\sqrt{3}}{2}$, on substituting this value in equation we will get,
 $\dfrac{\sqrt{3}}{2}=\dfrac{AB}{AC}$ ………………(i)
Or $AB=\dfrac{\sqrt{3}}{2}AC$
Now, we know that Pythagoras theorem for $\Delta ABC$ can be given as,
$A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
Now, on substituting the value of AB in expression we will get,
${{\left( \dfrac{\sqrt{3}}{2}AC \right)}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
On further simplifying the expression we will get,
$\dfrac{3}{4}A{{C}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
$\Rightarrow B{{C}^{2}}=A{{C}^{2}}-\dfrac{3}{4}A{{C}^{2}}$
Now, on taking LCM we will get,
$B{{C}^{2}}=\dfrac{4}{4}A{{C}^{2}}-\dfrac{3}{4}A{{C}^{2}}$
$\Rightarrow B{{C}^{2}}=\dfrac{4-3}{4}A{{C}^{2}}\Rightarrow B{{C}^{2}}=\dfrac{1}{4}A{{C}^{2}}$
$\Rightarrow \dfrac{B{{C}^{2}}}{A{{C}^{2}}}=\dfrac{1}{4}$
Now, on applying square roots on both the sides we will get,
$\Rightarrow \dfrac{BC}{AC}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$ ………………….(ii)
From expressions (i) and (ii), values of sides AB, BC and AC can be considered as $\sqrt{3}$, $1$ and $2$ respectively so the ratio of sides can be given as,
$AB:BC:AC=\sqrt{3}:1:2$ or $AC:AB:BC=2:\sqrt{3}:1$

Thus, option (c) is correct answer.

Note: Here, we calculated the problem considering the angle made by point C and using $\sin \theta =\dfrac{\text{Opposite}}{\text{Hypotenuse}}$, but one can also consider the cosine angle made by point C i.e. $\cos \theta =\dfrac{\text{Adjacent}}{\text{Hypotenuse}}$. By using this the sides will be $\cos 60=\dfrac{BC}{AC}=\dfrac{1}{2}$ and by this also we will get, values of sides BC and AC as 1 and 2 respectively and we will apply Pythagoras theorem then the value of AB will also be the same i.e. $\sqrt{3}$. So, we can consider this as an alternative approach to solve the problem.