Every problem has a full worked solution that names the angle relationship, writes the equation and checks the result, with the final answer in bold. Compare each step with your own work and redraw any figure you missed.
Problem 1: Find the third angle
The angles sum to \(180^{\circ}\): \(180-48-67=65\).
\(65^{\circ}\)
Problem 2: Right triangle angles
The right angle is \(90^{\circ}\), so \(180-90-35=55\).
\(55^{\circ}\)
Problem 3: Isosceles triangle
The base angles are equal and share \(180-40=140\) degrees, so each is \(140: 2=70\).
Each base angle is \(70^{\circ}\).
Problem 4: Angles in a ratio
Add and set equal to 180: \(x+2x+3x=6x=180\), so \(x=30\).
\(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\)
Problem 5: Exterior angle
An exterior angle equals the sum of the two remote interior angles: \(x=50+65=115\).
Check: the interior angle at \(B\) is \(180-50-65=65\), and \(65+115=180\).
\(x=115^{\circ}\)
Problem 6: Missing remote angle
The exterior angle is the sum of the remote angles: \(130=80+y\), so \(y=50\).
\(50^{\circ}\)
Problem 7: Corresponding angles
The marked angles are in the same position at each intersection, so they are corresponding angles. Parallel lines make them equal.
\(x=65^{\circ}\)
Problem 8: Alternate interior angles
The marked angles lie between the lines on opposite sides of the transversal: alternate interior angles. They are equal.
\(x=112^{\circ}\)
Problem 9: Same-side interior angles
The marked angles are between the lines on the same side of the transversal, so they add to \(180^{\circ}\): \(x=180-108=72\).
\(x=72^{\circ}\)
Problem 10: Linear pair and vertical angles
\(x\) and the \(55^{\circ}\) angle form a linear pair, so \(x=180-55=125\). \(y\) and the \(55^{\circ}\) angle are vertical angles, so \(y=55\).
\(x=125^{\circ}\) and \(y=55^{\circ}\)
Problem 11: Alternate exterior angles
The marked angles lie outside the lines on opposite sides of the transversal: alternate exterior angles. They are equal.
\(x=118^{\circ}\)
Problem 12: Corresponding angles with x
Corresponding angles are equal: \(3x+10=70\). Subtract 10: \(3x=60\). Divide by 3: \(x=20\).
Check: \(3(20)+10=70\).
\(x=20\)
Problem 13: Same-side interior with x
They add to 180: \((2x+20)+(x+40)=180\), so \(3x+60=180\) and \(3x=120\), giving \(x=40\).
Angles: \(2(40)+20=100\) and \(40+40=80\); \(100+80=180\).
\(x=40\); the angles are \(100^{\circ}\) and \(80^{\circ}\).
Problem 14: Triangle with x
\((x+10)+2x+(3x+20)=180\) gives \(6x+30=180\), so \(6x=150\) and \(x=25\).
Angles: \(35\), \(50\) and \(95\); sum \(=180\).
\(x=25\); angles \(35^{\circ}\), \(50^{\circ}\), \(95^{\circ}\)
Problem 15: Exterior angle with x
The exterior angle equals the sum of the remote angles: \(5x-10=(2x+15)+(x+25)=3x+40\).
Subtract \(3x\) and add 10: \(2x=50\), so \(x=25\). The exterior angle is \(5(25)-10=115\).
\(x=25\); exterior angle \(115^{\circ}\)
Problem 16: All eight angles
\(a\) forms a linear pair with the \(74^{\circ}\) angle: \(a=180-74=106\). \(c\) is vertical to \(a\), so \(c=106\). \(d\) is vertical to \(74^{\circ}\), so \(d=74\).
The angles at the bottom line match the top by corresponding angles: \(e=106\), \(f=74\), \(g=106\), \(h=74\).
\(a=c=e=g=106^{\circ}\) and \(d=f=h=74^{\circ}\)
Problem 17: Same angles?
Third angle of triangle 1: \(180-50-60=70\). Third angle of triangle 2: \(180-60-70=50\).
Both triangles have angles \(50^{\circ}\), \(60^{\circ}\) and \(70^{\circ}\).
Yes, they are similar by the AA criterion.
Problem 18: Not similar
Third angle of triangle 1: \(180-45-65=70\). Third angle of triangle 2: \(180-45-75=60\).
The angle sets \(45,65,70\) and \(45,75,60\) differ, even though one angle matches.
No, the triangles are not similar.
Problem 19: Missing side by AA
Angles \(A=D=60^{\circ}\) and \(B=E=70^{\circ}\), so \(\triangle ABC\sim\triangle DEF\) by AA.
Scale factor: \(DE: AB=9: 6=\frac{3}{2}\). Then \(EF=8\cdot\frac{3}{2}=12\).
\(EF=12\)
Problem 20: Roof truss
The three angles sum to \(180^{\circ}\): \(100+x+x=180\), so \(2x=80\) and \(x=40\).
Each base angle is \(40^{\circ}\).
Problem 21: City streets
East side: the two angles are same-side interior angles, so they add to \(180^{\circ}\): \(180-52=128\).
West side: this angle is an alternate interior angle to the \(52^{\circ}\) angle, so it equals \(52^{\circ}\). Check: \(128+52=180\), a linear pair.
East side \(128^{\circ}\); west side \(52^{\circ}\).
Problem 22: Parallel line through a vertex
\(x\) and \(\angle A\) are alternate interior angles, so \(x=58\). \(z\) and \(\angle B\) are alternate interior angles, so \(z=67\).
\(\angle C=180-58-67=55\). Along the straight line: \(58+55+67=180\).
\(x=58^{\circ}\), \(z=67^{\circ}\), \(\angle C=55^{\circ}\)
Problem 23: Ramp brace
The interior angle next to the exterior angle: \(180-125=55\).
The exterior angle equals the sum of the two remote angles: \(125=35+C\), so \(C=90\).
Check: \(55+35+90=180\).
\(55^{\circ}\), \(35^{\circ}\), \(90^{\circ}\)
Problem 24: Quilt seam
Alternate interior angles are equal: \(4x+8=6x-32\). Subtract \(4x\): \(8=2x-32\). Add 32: \(40=2x\), so \(x=20\).
Each angle: \(4(20)+8=88\) and \(6(20)-32=88\).
\(x=20\); each angle measures \(88^{\circ}\).
See every chapter on the Grade 8 math page.