91Ó°ÊÓ

We are given a right angled triangle and the measure of one small angle $18$less than twice the measure of the other small angle.

Let the small angles of the right triangle be $a,b$.

We know that the sum of angles of a triangle is equal to ${180}^{∘}$.

So, the equation will be $a+b+{90}^{∘}={180}^{∘}$

The one small angle is $18$less than twice the measure of the other small angle, so

$a=2b-18$

Substituting the value of $a$in first equation, we get

$$\begin{gathered} 2b-18+b+{90}^{∘}={180}^{∘} \\ 3b+{72}^{∘}={180}^{∘} \end{gathered}$$

Subtracting ${72}^{∘}$from both sides, we get

$3b={108}^{∘}$

Dividing by $3$,

$$\begin{gathered} b=\frac{{108}^{∘}}{3} \\ b={36}^{∘} \end{gathered}$$

Now, putting the value of $b$in second equation,

$$\begin{gathered} a=2×36-18 \\ ={54}^{∘} \end{gathered}$$

Checking the solution by putting the value of $a,b$in the equations, we get

$$\begin{gathered} a+b+{90}^{∘}={180}^{∘} \\ {54}^{∘}+{36}^{∘}+{90}^{∘}={180}^{∘} \\ {180}^{∘}={180}^{∘} \\ a=2b-18 \\ {54}^{∘}=2\left({36}^{∘}\right)-18 \\ {54}^{∘}={72}^{∘}-18 \\ {54}^{∘}={54}^{∘} \end{gathered}$$

This is true, hence the solution is correct.