 QUESTION

# In a right-angled triangle, one of the acute angles exceeds the other by ${{20}^{\circ }}$. Find the measure of both the acute angles in the right-angled triangle.a) ${{35}^{\circ }},{{55}^{\circ }}$b) ${{35}^{\circ }},{{25}^{\circ }}$c) ${{35}^{\circ }},{{35}^{\circ }}$d) ${{35}^{\circ }},{{75}^{\circ }}$

Hint: Let us assume that one of the angles of the right angled triangle other than right angle be x. Also, we know that, sum of all the angles of any triangle is ${{180}^{\circ }}$, which means that the sum of the angles other right angle of right angled triangle is ${{90}^{\circ }}$.

In this question, we have to find the angles of the right angled triangle, one of which exceeds the other by ${{20}^{\circ }}$ and they both are acute angles.
Let us consider that one of the angles of a right angled triangle be x, therefore, angle other than ${{90}^{\circ }}$ and x is ${{90}^{\circ }}-x$, because we know that sum of all angles of any triangle is ${{180}^{\circ }}$.
We have given that one of the acute angles exceeds the other by ${{20}^{\circ }}$. We can write it as $x={{90}^{\circ }}-x+{{20}^{\circ }}$. Here, we will take all terms of variable x on the left hand side and the rest of terms on the right hand side. So, we will get, $2x={{90}^{\circ }}+{{20}^{\circ }}$
$\Rightarrow 2x={{110}^{\circ }}$
And, on further simplification, we will get, $x={{55}^{\circ }}$
Therefore, we can see that one of the angles is ${{55}^{\circ }}$ and now, we will calculate other angle by subtracting first angle from ${{90}^{\circ }}$.
So, the second angle is ${{35}^{\circ }}$.
Note: We can also find the answer using the shortcut method. As we know sum of angles of right angled triangle other than right angle is ${{90}^{\circ }}$, and it is given that difference between the angles is of ${{20}^{\circ }}$. Now, we can use options to see which option satisfies both the condition. Here, we can see that only option (a) satisfies both the condition.