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 vertical 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 vertical 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° = 110°
x° = 22°
Thus, ∠AOB = 54°.

Solution:
The four angles add up to [tex]360^\circ[/tex], so the other three add up to [tex]360^\circ-\alpha[/tex]:
[tex]\alpha=\frac{2}{3}(360^\circ-\alpha) \Rightarrow 3\alpha=720^\circ-2\alpha \Rightarrow 5\alpha=720^\circ \Rightarrow \alpha=144^\circ[/tex]

Solution:
The two angles adjacent to [tex]\alpha[/tex] are equal (vertical angles), each [tex]180^\circ-\alpha[/tex]:
[tex]\alpha=4\times 2(180^\circ-\alpha) \Rightarrow 9\alpha=1440^\circ \Rightarrow \alpha=160^\circ[/tex]
Each adjacent angle is [tex]180^\circ-160^\circ=20^\circ[/tex].
[tex]50^\circ 38'-20^\circ=30^\circ 38'[/tex]

Solution:
[tex]\angle AOD=180^\circ-143^\circ=37^\circ[/tex]
[tex]OC[/tex] cannot lie between [tex]OA[/tex] and [tex]OD[/tex] (because [tex]37^\circ\lt 90^\circ[/tex]), so [tex]OD[/tex] lies between [tex]OA[/tex] and [tex]OC[/tex]:
[tex]\angle AOC=\angle AOD+\angle DOC=37^\circ+90^\circ=127^\circ[/tex]

Solution:
[tex]\angle DAM=2\times 38^\circ=76^\circ[/tex]
[tex]\angle CAM=180^\circ-76^\circ=104^\circ[/tex]
[tex]\angle LAM=\frac{104^\circ}{2}=52^\circ[/tex]

Solution:
[tex]5x+7x+2x=180^\circ \Rightarrow 14x=180^\circ[/tex]
[tex]\angle COD=7x=\frac{7\times 180^\circ}{14}=90^\circ[/tex]

Solution:
The bisectors of adjacent supplementary angles are perpendicular, so the angle between them is [tex]90^\circ[/tex].
Half of one of the angles is [tex]90^\circ-15^\circ=75^\circ[/tex], so this angle is [tex]150^\circ[/tex] and the other is [tex]30^\circ[/tex].

Solution:
Since [tex]\angle AOC\lt \angle BOC[/tex], the ray [tex]OA[/tex] lies inside [tex]\angle BOC[/tex]:
[tex]\angle BOC=\angle BOA+\angle AOC[/tex]
[tex]3x=50^\circ+x \Rightarrow 2x=50^\circ \Rightarrow x=25^\circ[/tex]

Solution:
Co-interior angles at parallel lines add up to [tex]180^\circ[/tex]:
[tex]x+(x+28^\circ)=180^\circ \Rightarrow x=76^\circ, x+28^\circ=104^\circ[/tex]
All eight angles are [tex]76^\circ[/tex] or [tex]104^\circ[/tex] (four of each).