Question

In the figure below, a triangle RST with $\angle S=90{}^\circ ,\angle T=45{}^\circ $, RT = 12 cm is given. Find RS and ST.

Answer 1

Hint: We will use that the sum of all three angles in a triangle is $180{}^\circ $ and we use Pythagoras theorem to find the length of sides as the above triangle is a right-angled triangle.

Complete step-by-step solution -
The sum of three angles in a triangle is $180{}^\circ $. So, we will use this condition to find the $\angle R$ in this triangle.
The above triangle is a right-angled triangle and right-angled at S.
So, $\angle S=90{}^\circ $ and $\angle T=45{}^\circ $. We will find $\angle R$ by subtracting $\angle S$ and $\angle T$ from $180{}^\circ $.
So, $\angle R=180{}^\circ -(\angle S+\angle T)$
$\begin{align}
  & \Rightarrow 180{}^\circ -(90{}^\circ +45{}^\circ ) \\
 & \Rightarrow 180{}^\circ -(135{}^\circ ) \\
 & \Rightarrow 45{}^\circ \\
\end{align}$
Both $\angle R$ and $\angle S$ are equal. So, the above triangle is a right-angled isosceles triangle.
In the above right-angled isosceles triangle, the hypotenuse is the side opposite to the right-angled triangle. So, the length of hypotenuse ST is 12 cm and let the length of side ST be x and length of the side RS also becomes x, as the triangle is an isosceles triangle.
We use Pythagoras theorem as the above triangle is a right-angled triangle, Pythagoras theorem states that in a right angle triangle square of the hypotenuse is equal to the sum of squares of other two sides in it.
So, by Pythagoras theorem $R{{S}^{2}}+S{{T}^{2}}=R{{T}^{2}}\cdot \cdot \cdot \cdot \left( 1 \right)$
Now we will substitute RT = 12 cm, RS = x cm, ST = x cm in equation (1).
Equation (1) becomes, $\Rightarrow {{x}^{2}}+{{x}^{2}}={{12}^{2}}$
 Now we will solve for x, $\Rightarrow 2{{x}^{2}}=144$
$\begin{align}
  & \Rightarrow {{x}^{2}}=72 \\
 & \Rightarrow x=\sqrt{72} \\
\end{align}$
$\sqrt{72}$ can be expressed as $\sqrt{36\times 2}$ then further simplifying it, we obtain $\pm 6\sqrt{2}$.
Therefore, $x=\pm 6\sqrt{2}$.
We do not consider $x=-6\sqrt{2}$ because the length of the side of a triangle cannot be negative.
Hence, $x=6\sqrt{2}$.
So, RS = ST = $6\sqrt{2}$.

Note: The length of the side of a triangle cannot be negative and in an isosceles triangle two sides will be equal. The Pythagora's theorem is only applicable to right-angled triangles. We can also solve this question using the relation between angles and sides of a right-angles triangle. We can compute side RS by using sine of angle $45{}^\circ $, as RS $=12\sin 45{}^\circ =6\sqrt{2}$ and ST using the cosine of angle $45{}^\circ $, as ST $=12\cos 45{}^\circ =6\sqrt{2}$.