Q 4.38 - Intermediate Algebra

Question

The measure of one of the small angles of a right triangle is $18$ less than twice the measure of the other small angle. Find the measure of both angles.

Step-by-Step Solution

Answer

The measure of small angles of right triangle is equal to ${54}^{\circ }$ and ${36}^{\circ }$.

Step 1 . Given

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.

Step 2 . Assumptions and formation of equation.

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

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

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

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

$a=2b-18$

Step 3 . Solving the equations

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

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

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

$3b={108}^{\circ }$

Dividing by $3$,

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

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

$$\begin{gathered} a=2\times 36-18 \\ ={54}^{\circ } \end{gathered}$$

Step 4. Checking the solution

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

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

This is true, hence the solution is correct.