Scroll:Geometry >> Angles >> ps (3113)

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)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=78°, ∠DBC=87° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




2)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=70°, ∠DBC=76° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




3)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=82°, ∠DBC=88° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




4)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=84°, ∠DBC=86° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




5)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=76°, ∠DBC=79° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




6)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=82°, ∠DBC=83° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




7)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straght lines. ∠DAF=78°, ∠DBC=88° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




8)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=74°, ∠DBC=82° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




9)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. ∠DAF=72°, ∠DBC=78° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




10)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines ∠DAF=72°, ∠DBC=77° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

°

b) ∠DCB

°


Answer:_______________




 

1)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=78°, ∠DBC=87° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 102°

b) ∠DCB

Answer: 54°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 78° (ED // FA)

                            = 102° 

 

b) Step 1: ∠DAB = 180° - 78° (sum of angles on a straight line)

                           = 102°

     Step 2:∠ABD +∠ADB =  180° - 102° (sum of angles of triangle)

                                        = 78°

     Step 3: ∠ABD = 78°÷ 2 (isosceles triangle)

                            = 39° 

     Step 4: ∠ABC = 87° + 39°

                            = 126°

     Step 5: ∠DCB = 180° - 126° (DC // AB)

                            = 54°



2)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=70°, ∠DBC=76° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 110°

b) ∠DCB

Answer: 69°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 70° (ED // FA)

                            = 110° 

 

b) Step 1: ∠DAB = 180° - 70° (sum of angles on a straight line)

                           = 110°

     Step 2:∠ABD +∠ADB =  180° - 110° (sum of angles of triangle)

                                        = 70°

     Step 3: ∠ABD = 70°÷ 2 (isosceles triangle)

                            = 35° 

     Step 4: ∠ABC = 76° + 35°

                            = 111°

     Step 5: ∠DCB = 180° - 111° (DC // AB)

                            = 69°



3)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=82°, ∠DBC=88° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 98°

b) ∠DCB

Answer: 51°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 82° (ED // FA)

                            = 98° 

 

b) Step 1: ∠DAB = 180° - 82° (sum of angles on a straight line)

                           = 98°

     Step 2:∠ABD +∠ADB =  180° - 98° (sum of angles of triangle)

                                        = 82°

     Step 3: ∠ABD = 82°÷ 2 (isosceles triangle)

                            = 41° 

     Step 4: ∠ABC = 88° + 41°

                            = 129°

     Step 5: ∠DCB = 180° - 129° (DC // AB)

                            = 41°



4)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=84°, ∠DBC=86° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 96°

b) ∠DCB

Answer: 52°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 84° (ED // FA)

                            = 96° 

 

b) Step 1: ∠DAB = 180° - 84° (sum of angles on a straight line)

                           = 96°

     Step 2:∠ABD +∠ADB =  180° - 96° (sum of angles of triangle)

                                        = 84°

     Step 3: ∠ABD = 84°÷ 2 (isosceles triangle)

                            = 42° 

     Step 4: ∠ABC = 86° + 42°

                            = 128°

     Step 5: ∠DCB = 180° - 128° (DC // AB)

                            = 52°



5)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=76°, ∠DBC=79° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 104°

b) ∠DCB

Answer: 63°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 76° (ED // FA)

                            = 104° 

 

b) Step 1: ∠DAB = 180° - 76° (sum of angles on a straight line)

                           = 104°

     Step 2:∠ABD +∠ADB =  180° - 104° (sum of angles of triangle)

                                        = 76°

     Step 3: ∠ABD = 76°÷ 2 (isosceles triangle)

                            = 38° 

     Step 4: ∠ABC = 79° + 38°

                            = 117°

     Step 5: ∠DCB = 180° - 117° (DC // AB)

                            = 63°



6)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=82°, ∠DBC=83° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 98°

b) ∠DCB

Answer: 56°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 82° (ED // FA)

                            = 98° 

 

b) Step 1: ∠DAB = 180° - 82° (sum of angles on a straight line)

                           = 98°

     Step 2:∠ABD +∠ADB =  180° - 98° (sum of angles of triangle)

                                        = 82°

     Step 3: ∠ABD = 82°÷ 2 (isosceles triangle)

                            = 41° 

     Step 4: ∠ABC = 83° + 41°

                            = 124°

     Step 5: ∠DCB = 180° - 124° (DC // AB)

                            = 56°



7)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straght lines. ∠DAF=78°, ∠DBC=88° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 102°

b) ∠DCB

Answer: 53°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 78° (ED // FA)

                            = 102° 

 

b) Step 1: ∠DAB = 180° - 78° (sum of angles on a straight line)

                           = 102°

     Step 2:∠ABD +∠ADB =  180° - 102° (sum of angles of triangle)

                                        = 78°

     Step 3: ∠ABD = 78°÷ 2 (isosceles triangle)

                            = 39° 

     Step 4: ∠ABC = 88° + 39°

                            = 127°

     Step 5: ∠DCB = 180° - 127° (DC // AB)

                            = 53°



8)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines. ∠DAF=74°, ∠DBC=82° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 106°

b) ∠DCB

Answer: 61°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 74° (ED // FA)

                            = 106° 

 

b) Step 1: ∠DAB = 180° - 74° (sum of angles on a straight line)

                           = 106°

     Step 2:∠ABD +∠ADB =  180° - 106° (sum of angles of triangle)

                                        = 74°

     Step 3: ∠ABD = 74°÷ 2 (isosceles triangle)

                            = 37° 

     Step 4: ∠ABC = 82° + 37°

                            = 119°

     Step 5: ∠DCB = 180° - 119° (DC // AB)

                            = 61°



9)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. ∠DAF=72°, ∠DBC=78° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 108°

b) ∠DCB

Answer: 66°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 72° (ED // FA)

                            = 108° 

 

b) Step 1: ∠DAB = 180° - 72° (sum of angles on a straight line)

                           = 108°

     Step 2:∠ABD +∠ADB =  180° - 108° (sum of angles of triangle)

                                        = 72°

     Step 3: ∠ABD = 72°÷ 2 (isosceles triangle)

                            = 36° 

     Step 4: ∠ABC = 78° + 36°

                            = 114°

     Step 5: ∠DCB = 180° - 114° (DC // AB)

                            = 66°



10)  

Given that ABCD is a trapezium and ABD is an isosceles triangle. EDC and FAB are straight lines ∠DAF=72°, ∠DBC=77° and BC=CD.

EC is parallel to FB. Calculate.

a) ∠EDA

Answer: 108°

b) ∠DCB

Answer: 67°


SOLUTION 1 :

a) Step 1: ∠EDA = 180° - 72° (ED // FA)

                            = 108° 

 

b) Step 1: ∠DAB = 180° - 72° (sum of angles on a straight line)

                           = 108°

     Step 2:∠ABD +∠ADB =  180° - 108° (sum of angles of triangle)

                                        = 72°

     Step 3: ∠ABD = 72°÷ 2 (isosceles triangle)

                            = 36° 

     Step 4: ∠ABC = 77° + 36°

                            = 113°

     Step 5: ∠DCB = 180° - 113° (DC // AB)

                            = 67°