Q 4.37 - Intermediate Algebra

Question

The measure of one of the small angles of a right triangle is $2$ more than $3$ times 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 ${68}^{\circ }$ and ${22}^{\circ }$.

Step 1 . Given

We are given a right-angled triangle and the measure of one small angle is $2$ more than $3$ times 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 $2$ more than $3$ times the measure of the other small angle, So,

$a=3b+2$

Step 3 . Solving the equations

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

$$\begin{gathered} 3b+2+b+{90}^{\circ }={180}^{\circ } \\ 4b+{92}^{\circ }={180}^{\circ } \end{gathered}$$

Subtracting ${92}^{\circ }$ from both sides, we get

$4b={88}^{\circ }$

Dividing by $4$, we get

$$\begin{gathered} b=\frac{{88}^{\circ }}{4} \\ b={22}^{\circ } \end{gathered}$$

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

$$\begin{gathered} a=3\times 22+2 \\ ={68}^{\circ } \end{gathered}$$

Step 4. Checking the solution

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

$$\begin{gathered} a+b+{90}^{\circ }={180}^{\circ } \\ {68}^{\circ }+{22}^{\circ }+{90}^{\circ }={180}^{\circ } \\ {180}^{\circ }={180}^{\circ } \\ a=3b+2 \\ {68}^{\circ }=3\left({22}^{\circ }\right)+2 \\ {68}^{\circ }={66}^{\circ }+2 \\ {68}^{\circ }={68}^{\circ } \end{gathered}$$

This is true, hence the solution is correct.