Angles: Difficult Problems with Solutions

Solution:
Since the angles are adjacent, they share vertex B and side BC. ∠ABD=∠ABC+∠DBC=53°+29°=82°.

Solution:
∠ABC and ∠CBD are adjacent angles. Thus, ∠ABC=∠ABD-∠DBC=127°-85°=42°.

Solution:
Since AC is the bisector of ∠BAD, this angle is two times greater than ∠BAC. The measure of ∠BAD is 104°.

Solution:
If ∠AOB and ∠BOC are adjacent and complementary, then
∠AOB + ∠BOC = 90°
x° + 8° + 2x° -17° = 90°
3x° - 9° = 90°
3x° = 99°
x° = 33°
Thus, ∠AOB = 41°

Solution:
OE is the bisector, then ∠BOC = 2∠EOC = 60°. ∠AOB and ∠BOC are supplementary, so ∠AOB = 180° - 60° = 120°.

Solution:
∠13 and ∠11 are alternate interior angles. Since lines a and b are parallel, the measures of these angles are equal. Thus, ∠11 also has 40°. ∠11 and ∠3 are corresponding. Since lines c and d are parallel, the measures of these angles are equal. Thus, ∠3 has ∠40°. ∠3 and ∠2 are adjacent and supplementary. Thus, ∠2 = 180° - 40° = 140°.

Solution:
∠AOC and ∠BOC are adjacent and supplementary. Thus, ∠AOC + ∠BOC = 180°.
x° + 4x° = 180°
5x° = 180°
x°=36°
Thus, m∠AOC = 36° and m∠BOC = 144°. ∠AOD and ∠BOC are opposite angles, so m∠AOD = 144°.

Solution:
∠AOC and ∠BOC are adjacent and supplementary. Thus, ∠AOC + ∠BOC = 180°.
x + 20° + x = 180°
2x + 20° = 180°
2x = 160°
x = 80°
Thus, m∠BOC = 80° and m∠AOC = 100°. ∠BOD and ∠AOC are opposite angles, so its measure is 100°.

Solution:
Since they are supplementary, their sum is 180°.
2x° + 10° + 80° + 3x° -20° =180°
5x° + 70° = 180°
5x° = 120°
x° = 22°
Thus, ∠AOB = 54°.