Question

1. in $\triangle rst$, if $overline{rt}congoverline{st}$, $mangle r=(9x + 2)^{circ}$, $mangle s=(13x - 18)^{circ}$, and $mangle t=(17x+1)^{circ}$, find $x$ and the measure of each angle.
2. in $\triangle def$, if $angle dcongangle e$, $de = x + 4$, $ef = 4x - 8$, and $df = 7x - 35$, find $x$ and the measure of each side.
3. in $\triangle wxy$, if $overline{wx}congoverline{wy}$, $mangle w=(x + 6)^{circ}$, $mangle x=(5x - 12)^{circ}$, and $mangle y=(7x - 48)^{circ}$, find $x$ and the measure of each angle.

equilateral triangles
a triangle is equilateral if and only if it is equiangular!

• if $mangle a=mangle b=mangle c$, then $overline{ab}=overline{bc}=overline{ac}$
• if $ab = bc = ac$, then $angle a=angle b=angle c$

find each missing measure.
11.
12.

1. if $\triangle pqr$ is an equilateral triangle, solve for $x$ and $y$.

Explanation:

Step1: Use angle - sum property of a triangle

In a triangle, the sum of interior angles is 180°. For \(\triangle RST\) with \(\angle R=(9x + 2)^{\circ}\), \(\angle S=(13x-18)^{\circ}\), and \(\angle T=(17x + 1)^{\circ}\), we have \((9x + 2)+(13x-18)+(17x + 1)=180\).
Combining like - terms: \((9x+13x + 17x)+(2-18 + 1)=180\), which simplifies to \(39x-15 = 180\).
Adding 15 to both sides: \(39x=180 + 15=195\).
Dividing both sides by 39: \(x = 5\).

Step2: Find the measure of each angle

\(m\angle R=9x + 2=9\times5+2=45 + 2=47^{\circ}\).
\(m\angle S=13x-18=13\times5-18=65-18 = 47^{\circ}\).
\(m\angle T=17x + 1=17\times5+1=85 + 1=86^{\circ}\).

For \(\triangle DEF\) with \(\angle D\cong\angle E\), so \(DF = EF\).
We set up the equation \(7x-35=4x-8\).
Subtracting \(4x\) from both sides: \(7x-4x-35=4x-4x-8\), which gives \(3x-35=-8\).
Adding 35 to both sides: \(3x=-8 + 35=27\).
Dividing both sides by 3: \(x = 9\).
\(DE=x + 4=9 + 4=13\).
\(EF=4x-8=4\times9-8=36-8 = 28\).
\(DF=7x-35=7\times9-35=63-35 = 28\).

For \(\triangle WXY\) with \(\overline{WX}\cong\overline{WY}\), so \(\angle X=\angle Y\).
We set up the equation \(5x-12=7x-48\).
Subtracting \(5x\) from both sides: \(5x-5x-12=7x-5x-48\), which gives \(-12=2x-48\).
Adding 48 to both sides: \(-12 + 48=2x-48 + 48\), so \(36=2x\).
Dividing both sides by 2: \(x = 18\).
\(m\angle W=x + 6=18+6=24^{\circ}\).
\(m\angle X=5x-12=5\times18-12=90 - 12=78^{\circ}\).
\(m\angle Y=7x-48=7\times18-48=126-48=78^{\circ}\).

For \(\triangle PQR\) which is equilateral, all angles are 60°.
We have the equation \(7x-3=60\).
Adding 3 to both sides: \(7x=60 + 3=63\).
Dividing both sides by 7: \(x = 9\).
Also, \(14y-59=9y + 1\).
Subtracting \(9y\) from both sides: \(14y-9y-59=9y-9y + 1\), which gives \(5y-59=1\).
Adding 59 to both sides: \(5y=1 + 59=60\).
Dividing both sides by 5: \(y = 12\).

For \(\triangle JKL\) which is equilateral, \(m\angle J=m\angle K=m\angle L = 60^{\circ}\).

Answer:

For \(\triangle RST\): \(x = 5\), \(m\angle R=47^{\circ}\), \(m\angle S=47^{\circ}\), \(m\angle T=86^{\circ}\).
For \(\triangle DEF\): \(x = 9\), \(DE = 13\), \(EF = 28\), \(DF = 28\).
For \(\triangle WXY\): \(x = 18\), \(m\angle W=24^{\circ}\), \(m\angle X=78^{\circ}\), \(m\angle Y=78^{\circ}\).
For \(\triangle PQR\): \(x = 9\), \(y = 12\).
For \(\triangle JKL\): \(m\angle J=60^{\circ}\), \(m\angle K=60^{\circ}\), \(m\angle L=60^{\circ}\).