Question

angle sum theorem in triangle worksheet
find the value of each unknown angle.
1
2
3
4
5
6
7

Explanation:

Step1: Use the angle - sum property of a triangle

The sum of angles in a triangle is \(180^{\circ}\). For example, in a right - angled triangle (if applicable), one angle is \(90^{\circ}\), and we can find the other angles using \(180^{\circ}-\text{(sum of known angles)}\). Also, vertical angles are equal. For example, \(\angle x=\angle y\) if they are vertical angles formed by two intersecting lines.

Step2: Calculate for each problem

Problem 1:

In \(\triangle ABE\), \(\angle B = 47^{\circ}\), \(\angle BAE=35^{\circ}\). Using the angle - sum property of a triangle \(\angle AEB=180-(47 + 35)=98^{\circ}\). Since \(\angle AEB+\angle AEC = 180^{\circ}\) (linear pair), \(\angle AEC=180 - 98=82^{\circ}\)

Problem 2:

In right - triangle \(PQR\) (\(\angle Q = 90^{\circ}\)), \(\angle QPR=24^{\circ}\). Then \(\angle PRQ=180-(90 + 24)=66^{\circ}\). In \(\triangle PRS\), \(\angle RPS=24 + 35=59^{\circ}\), \(\angle PRS = 66^{\circ}\). Using the angle - sum property of a triangle, \(\angle PSR=180-(59+66)=55^{\circ}\)

Problem 3:

In \(\triangle PRS\), \(\angle P = 66^{\circ}\), \(\angle R = 72^{\circ}\). Using the angle - sum property of a triangle, \(\angle RSP=180-(66 + 72)=42^{\circ}\). Since \(\angle RSP\) and \(\angle QSR\) are vertical angles, \(\angle QSR = 42^{\circ}\)

Problem 4:

In right - triangle \(POS\) (\(\angle P = 90^{\circ}\)), \(\angle S = 46^{\circ}\). Then \(\angle PQS=180-(90 + 46)=44^{\circ}\). \(\angle RQS=180 - 44=136^{\circ}\)

Problem 5:

In \(\triangle SQT\), \(\angle S = 50^{\circ}\), \(\angle T = 80^{\circ}\). Then \(\angle SQT=180-(50 + 80)=50^{\circ}\). \(\angle QRT\) is an exterior angle of \(\triangle PQT\). \(\angle QRT=\angle P+\angle PTQ\). First, in \(\triangle PQT\), \(\angle PTQ = 180-(90 + 65)=25^{\circ}\). So \(\angle QRT=65 + 25=90^{\circ}\)

Problem 6:

\(\angle PQU\) and \(\angle SQU\) are supplementary. \(\angle PQU = 180-(x + 74)\). Also, using the angle - sum property of \(\triangle QRS\) and vertical angles (\(\angle QSU = 33^{\circ}\)). \(\angle RQS=33^{\circ}\). Then \(180-(x + 74)=33+(38 + x)\). Solving \(180-x - 74=33+38+x\), \(106-x=71+x\), \(2x=35\), \(x = 17.5\). \(\angle PQU=180-(17.5+74)=88.5^{\circ}\), \(\angle QRS=38 + 17.5=55.5^{\circ}\)

Problem 7:

In \(\triangle CDE\), \(\angle D = 43^{\circ}\), \(\angle E = 67^{\circ}\). Then \(\angle DCE=180-(43 + 67)=70^{\circ}\). Since \(\angle ACB=\angle DCE = 70^{\circ}\) (vertical angles). In \(\triangle ABC\), \(\angle A=\angle B=x\). Using the angle - sum property \(x + x+70=180\), \(2x=110\), \(x = 55^{\circ}\). So \(\angle ABC = 55^{\circ}\), \(\angle BAC=55^{\circ}\)

Answer:

1. \(\angle AEB = 98^{\circ}\), \(\angle AEC=82^{\circ}\)
2. \(\angle PRQ = 66^{\circ}\), \(\angle PSR=55^{\circ}\)
3. \(\angle RSP = 42^{\circ}\), \(\angle QSR=42^{\circ}\)
4. \(\angle RQS = 136^{\circ}\)
5. \(\angle QRT = 90^{\circ}\)
6. \(\angle PQU = 88.5^{\circ}\), \(\angle QRS=55.5^{\circ}\)
7. \(\angle ABC = 55^{\circ}\), \(\angle BAC=55^{\circ}\)