Question

1. find the values of x and y in the diagram below.
2. in △mnp, if m∠m=(4x - 3)°, m∠n=(9x - 6)°, and m∠p=(6x - 1)°, find the value of x and the measure of each angle.
3. in △rst, if m∠r is five more than twice x, m∠s is one more than x, and m∠t is sixteen less than seven times x, find x and the measure of each angle.
4. in △abc, if m∠a is thirteen less than m∠c and m∠b is eleven less than four times m∠c, find the measure of each angle.

in △jkl, if m∠k is nine more than m∠j and m∠l is 21 less than twice m∠j, find the measure of each angle.

Explanation:

Step1: Recall angle - sum property of a triangle

The sum of the interior angles of a triangle is 180 degrees.

Step2: Solve problem 13

In \(\triangle MNP\), we have \(m\angle M=(4x - 3)^{\circ}\), \(m\angle N=(2x - 6)^{\circ}\), and \(m\angle P=(6x - 1)^{\circ}\).
By the angle - sum property of a triangle: \((4x - 3)+(2x - 6)+(6x - 1)=180\).
Combine like terms: \(4x+2x + 6x-3-6 - 1=180\), which simplifies to \(12x-10 = 180\).
Add 10 to both sides: \(12x=190\), then \(x=\frac{190}{12}=\frac{95}{6}\approx15.83\).
\(m\angle M=4x - 3=4\times\frac{95}{6}-3=\frac{380}{6}-3=\frac{380 - 18}{6}=\frac{362}{6}=\frac{181}{3}\approx60.33^{\circ}\).
\(m\angle N=2x - 6=2\times\frac{95}{6}-6=\frac{190}{6}-6=\frac{190 - 36}{6}=\frac{154}{6}=\frac{77}{3}\approx25.67^{\circ}\).
\(m\angle P=6x - 1=6\times\frac{95}{6}-1=95 - 1 = 94^{\circ}\).

Step3: Solve problem 14

In \(\triangle RST\), \(m\angle R = 2x+5\), \(m\angle S=x + 1\), \(m\angle T=7x-16\).
By the angle - sum property: \((2x + 5)+(x + 1)+(7x-16)=180\).
Combine like terms: \(2x+x + 7x+5 + 1-16=180\), which simplifies to \(10x-10 = 180\).
Add 10 to both sides: \(10x=190\), so \(x = 19\).
\(m\angle R=2x+5=2\times19+5=38 + 5=43^{\circ}\).
\(m\angle S=x + 1=19 + 1=20^{\circ}\).
\(m\angle T=7x-16=7\times19-16=133-16 = 117^{\circ}\).

Step4: Solve problem 15

In \(\triangle ABC\), let \(m\angle C=x\). Then \(m\angle A=x - 13\) and \(m\angle B=4x-11\).
By the angle - sum property: \((x - 13)+(4x-11)+x=180\).
Combine like terms: \(x+4x + x-13-11=180\), which simplifies to \(6x-24 = 180\).
Add 24 to both sides: \(6x=204\), so \(x = 34\).
\(m\angle A=x - 13=34-13 = 21^{\circ}\).
\(m\angle B=4x-11=4\times34-11=136-11 = 125^{\circ}\).
\(m\angle C=34^{\circ}\).

Step5: Solve problem 16

In \(\triangle JKL\), let \(m\angle J=x\). Then \(m\angle K=x + 9\) and \(m\angle L=2x-21\).
By the angle - sum property: \(x+(x + 9)+(2x-21)=180\).
Combine like terms: \(x+x + 2x+9-21=180\), which simplifies to \(4x-12 = 180\).
Add 12 to both sides: \(4x=192\), so \(x = 48\).
\(m\angle J=48^{\circ}\).
\(m\angle K=x + 9=48+9 = 57^{\circ}\).
\(m\angle L=2x-21=2\times48-21=96-21 = 75^{\circ}\).

Answer:

Problem 13: \(x=\frac{95}{6}\), \(m\angle M=\frac{181}{3}^{\circ}\), \(m\angle N=\frac{77}{3}^{\circ}\), \(m\angle P = 94^{\circ}\)
Problem 14: \(x = 19\), \(m\angle R=43^{\circ}\), \(m\angle S=20^{\circ}\), \(m\angle T=117^{\circ}\)
Problem 15: \(x = 34\), \(m\angle A=21^{\circ}\), \(m\angle B=125^{\circ}\), \(m\angle C=34^{\circ}\)
Problem 16: \(x = 48\), \(m\angle J=48^{\circ}\), \(m\angle K=57^{\circ}\), \(m\angle L=75^{\circ}\)