Question

use the angle addition postulate to find the value of x and the measure of the angle indicated for each of the following problems.

1. m∠sxt=(4x + 1)°, m∠qxs=(2x - 2)°, and m∠qxt = 125°. find the value of x and m∠qxs.
2. m∠rxq=(x + 15)°, m∠rxs=(5x - 7)°, and m∠qxs=(3x+5)°. find the value of x and m∠rxs.

Explanation:

Step1: Apply angle - addition postulate

According to the angle - addition postulate, \(m\angle SXT+m\angle QXS = m\angle QXT\). Substitute the given angle measures: \((4x + 1)+(2x-2)=125\).

Step2: Simplify the left - hand side

Combine like terms: \(4x+2x + 1-2=125\), which simplifies to \(6x - 1=125\).

Step3: Solve for \(x\)

Add 1 to both sides: \(6x=125 + 1=126\). Then divide both sides by 6: \(x=\frac{126}{6}=21\).

Step4: Find \(m\angle QXS\)

Substitute \(x = 21\) into the expression for \(m\angle QXS\): \(m\angle QXS=(2x-2)^{\circ}=(2\times21 - 2)^{\circ}=(42-2)^{\circ}=40^{\circ}\).

for 13:

Step1: Apply angle - addition postulate

Since \(m\angle RXQ+m\angle QXS=m\angle RXS\), substitute the given angle measures: \((x + 15)+(3x + 5)=5x-7\).

Step2: Simplify the left - hand side

Combine like terms: \(x+3x+15 + 5=5x-7\), which gives \(4x+20=5x-7\).

Step3: Solve for \(x\)

Subtract \(4x\) from both sides: \(20=x - 7\). Then add 7 to both sides: \(x=20 + 7=27\).

Step4: Find \(m\angle RXS\)

Substitute \(x = 27\) into the expression for \(m\angle RXS\): \(m\angle RXS=(5x-7)^{\circ}=(5\times27-7)^{\circ}=(135 - 7)^{\circ}=128^{\circ}\).

Answer:

\(x = 21\)
\(m\angle QXS=40^{\circ}\)