Scroll:Geometry >> Angles >> ps (3083)
Written Instructions:
For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..
For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.
Leave your answers in the simplest form or correct to two decimal places.
| 1) ∠CDE=77°, ∠BCD=124° and ∠CBA=137° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 2) ∠CDE=62°, ∠BCD=124° and ∠CBA=141° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 3) ∠CDE=78°, ∠BCD=124° and ∠CBA=135° and BC=CD. DEA is a straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 4) ∠CDE=84°, ∠BCD=122° and ∠CBA=140° and BC=CD. DEA is a straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 5) ∠CDE=81°, ∠BCD=126° and ∠CBA=142° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 6) ∠CDE=77°, ∠BCD=120° and ∠CBA=131° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 7) ∠CDE=65°, ∠BCD=112° and ∠CBA=138° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 8) ∠CDE=83°, ∠BCD=114° and ∠CBA=143° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 9) ∠CDE=79°, ∠BCD=120° and ∠CBA=141° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 10) ∠CDE=82°, ∠BCD=112° and ∠CBA=141° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE ° b) ∠BAE °
Answer:_______________ |
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| 1) ∠CDE=77°, ∠BCD=124° and ∠CBA=137° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 49° b) ∠BAE Answer: 22°
SOLUTION 1 : a) Step 1: 180° - 124° = 56° (sum of angles of triangle) Step 2: ∠CDB = 56° ÷ 2 (isosceles triangle) = 28° Step 3: ∠BDE = 77° - 28° = 49°
(b) Step 1: ∠DEB = 180° - 77° (BE // CD) = 103° Step 2: ∠BEA = 180° - 103° (sum of angles on a straight line) = 77° Step 3: ∠EBC = 180° - 124° (DE // CB) = 56° Step 4: ∠EBA = 137° - 56° = 81° Step 5: ∠BAE = 180° - 77° - 81° (sum of angles of triangle) = 22° |
| 2) ∠CDE=62°, ∠BCD=124° and ∠CBA=141° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 34° b) ∠BAE Answer: 33°
SOLUTION 1 : a) Step 1: 180° - 124° = 56° (sum of angles of triangle) Step 2: ∠CDB = 56° ÷ 2 (isosceles triangle) = 28° Step 3: ∠BDE = 62° - 28° = 34°
(b) Step 1: ∠DEB = 180° - 62° (BE // CD) = 118° Step 2: ∠BEA = 180° - 118° (sum of angles on a straight line) = 62° Step 3: ∠EBC = 180° - 124° (DE // CB) = 56° Step 4: ∠EBA = 141° - 56° = 85° Step 5: ∠BAE = 180° - 85° - 62° (sum of angles of triangle) = 33° |
| 3) ∠CDE=78°, ∠BCD=124° and ∠CBA=135° and BC=CD. DEA is a straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 50° b) ∠BAE Answer: 23°
SOLUTION 1 : a) Step 1: 180° - 124° = 56° (sum of angles of triangle) Step 2: ∠CDB = 56° ÷ 2 (isosceles triangle) = 28° Step 3: ∠BDE = 78° - 28° = 50°
(b) Step 1: ∠DEB = 180° - 78° (BE // CD) = 102° Step 2: ∠BEA = 180° - 102° (sum of angles on a straight line) = 78° Step 3: ∠EBC = 180° - 124° (DE // CB) = 56° Step 4: ∠EBA = 130° - 56° = 79° Step 5: ∠BAE = 180° - 79° - 78° (sum of angles of triangle) = 23° |
| 4) ∠CDE=84°, ∠BCD=122° and ∠CBA=140° and BC=CD. DEA is a straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 55° b) ∠BAE Answer: 14°
SOLUTION 1 : a) Step 1: 180° - 122° = 58° (sum of angles of triangle) Step 2: ∠CDB = 58° ÷ 2 (isosceles triangle) = 29° Step 3: ∠BDE = 84° - 29° = 55°
(b) Step 1: ∠DEB = 180° - 84° (BE // CD) = 96° Step 2: ∠BEA = 180° - 96° (sum of angles on a straight line) = 84° Step 3: ∠EBC = 180° - 122° (DE // CB) = 58° Step 4: ∠EBA = 140° - 58° = 82° Step 5: ∠BAE = 180° - 84° - 82° (sum of angles of triangle) = 14° |
| 5) ∠CDE=81°, ∠BCD=126° and ∠CBA=142° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 54° b) ∠BAE Answer: 11°
SOLUTION 1 : a) Step 1: 180° - 126° = 54° (sum of angles of triangle) Step 2: ∠CDB = 54° ÷ 2 (isosceles triangle) = 27° Step 3: ∠BDE = 81° - 27° = 54°
(b) Step 1: ∠DEB = 180° - 81° (BE // CD) = 99° Step 2: ∠BEA = 180° - 99° (sum of angles on a straight line) = 81° Step 3: ∠EBC = 180° - 126° (DE // CB) = 54° Step 4: ∠EBA = 142° - 54° = 88° Step 5: ∠BAE = 180° - 81° - 88° (sum of angles of triangle) = 11° |
| 6) ∠CDE=77°, ∠BCD=120° and ∠CBA=131° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 47° b) ∠BAE Answer: 32°
SOLUTION 1 : a) Step 1: 180° - 120° = 60° (sum of angles of triangle) Step 2: ∠CDB = 60° ÷ 2 (isosceles triangle) = 30° Step 3: ∠BDE = 77° - 30° = 47°
(b) Step 1: ∠DEB = 180° - 77° (BE // CD) = 103° Step 2: ∠BEA = 180° - 103° (sum of angles on a straight line) = 77° Step 3: ∠EBC = 180° - 120° (DE // CB) = 60° Step 4: ∠EBA = 131° - 60° = 71° Step 5: ∠BAE = 180° - 71° - 77° (sum of angles of triangle) = 32° |
| 7) ∠CDE=65°, ∠BCD=112° and ∠CBA=138° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 31° b) ∠BAE Answer: 45°
SOLUTION 1 : a) Step 1: 180° - 112° = 68° (sum of angles of triangle) Step 2: ∠CDB = 68° ÷ 2 (isosceles triangle) = 34° Step 3: ∠BDE = 65° - 34° = 31°
(b) Step 1: ∠DEB = 180° - 65° (BE // CD) = 115° Step 2: ∠BEA = 180° - 115° (sum of angles on a straight line) = 65° Step 3: ∠EBC = 180° - 112° (DE // CB) = 68° Step 4: ∠EBA = 138° - 68° = 70° Step 5: ∠BAE = 180° - 70° - 65° (sum of angles of triangle) = 45° |
| 8) ∠CDE=83°, ∠BCD=114° and ∠CBA=143° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 50° b) ∠BAE Answer: 20°
SOLUTION 1 : a) Step 1: 180° - 114° = 66° (sum of angles of triangle) Step 2: ∠CDB = 66° ÷ 2 (isosceles triangle) = 33° Step 3: ∠BDE = 83° - 33° = 50°
(b) Step 1: ∠DEB = 180° - 83° (BE // CD) = 97° Step 2: ∠BEA = 180° - 97° (sum of angles on a straight line) = 83° Step 3: ∠EBC = 180° - 114° (DE // CB) = 66° Step 4: ∠EBA = 143° - 66° = 77° Step 5: ∠BAE = 180° - 83° - 77° (sum of angles of triangle) = 20° |
| 9) ∠CDE=79°, ∠BCD=120° and ∠CBA=141° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 49° b) ∠BAE Answer: 20°
SOLUTION 1 : a) Step 1: 180° - 120° = 60° (sum of angles of triangle) Step 2: ∠CDB = 60° ÷ 2 (isosceles triangle) = 30° Step 3: ∠BDE = 79° - 30° = 49°
(b) Step 1: ∠DEB = 180° - 79° (BE // CD) = 101° Step 2: ∠BEA = 180° - 101° (sum of angles on a straight line) = 79° Step 3: ∠EBC = 180° - 120° (DE // CB) = 60° Step 4: ∠EBA = 141° - 60° = 81° Step 5: ∠BAE = 180° - 81° - 79° (sum of angles of triangle) = 20° |
| 10) ∠CDE=82°, ∠BCD=112° and ∠CBA=141° and BC=CD. DEA is straight line. The point E on AD is such that BE is parallel to CD. Calculate. a) ∠BDE Answer: 48° b) ∠BAE Answer: 25°
SOLUTION 1 : a) Step 1: 180° - 112° = 68° (sum of angles of triangle) Step 2: ∠CDB = 68° ÷ 2 (isosceles triangle) = 34° Step 3: ∠BDE = 82° - 34° = 48°
(b) Step 1: ∠DEB = 180° - 82° (BE // CD) = 92° Step 2: ∠BEA = 180° - 92° (sum of angles on a straight line) = 92° Step 3: ∠EBC = 180° - 112° (DE // CB) = 68° Step 4: ∠EBA = 141° - 68° = 73° Step 5: ∠BAE = 180° - 73° - 82° (sum of angles of triangle) = 25° |