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A Question For The Mathematicians In The House by Timijo(m): 5:35am On Feb 21, 2022
I have tried to apply all the angle theorems that I know to answer question b but my result does not correspond to the rules.
Can we assume that angles A and C are 90⁰?

Note: AD is not parallel to BC.
Look at the diagram below, if you know the right theorem to use, please solve it
Thanks in anticipation.

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Re: A Question For The Mathematicians In The House by dejt4u(m): 6:16am On Feb 21, 2022
Timijo:
I have tried to apply all the angle theorems that I know to answer question b but my result does not correspond to the rules.
Can we assume that angles A and C are 90⁰?

Note: AD is not parallel to BC.
Look at the diagram below, if you know the right theorem to use, please solve it
Thanks in anticipation.
very simple question.

Split the shape into two triangles. If the angle at the middle is O.
Then we have triangles ADO and BCO.

But <C is 60° (i.e, (360 - (120×2))/2),

Therefore, A + D + 60° = 180°
A + D = 120°.......eqn 1
Similarly, B + C + 60° = 180°
B + C = 120° ........eqn 2

(a) sum of angles A, B, C and D can be established by adding eqn 1 and eqn 2 together. That is,
A + D + B + C = 120 + 120,
A + B + C + D = 240°.
therefore, the sum of angles A, B, C and D is 240° which is less than 360°


(b) A + B + C + D = 240°.

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Re: A Question For The Mathematicians In The House by Timijo(m): 7:26am On Feb 21, 2022
dejt4u:
very simple question.

Split the shape into two triangles. If the angle at the middle is O.
Then we have triangles ADO and BCO.

But <C is 60° (i.e, (360 - (120×2))/2),

Therefore, A + D + 60° = 180°
A + D = 120°.......eqn 1
Similarly, B + C + 60° = 180°
B + C = 120° ........eqn 2

(a) sum of angles A, B, C and D can be established by adding eqn 1 and eqn 2 together. That is,
A + D + B + C = 120 + 120,
A + B + C + D = 240°.
therefore, the sum of angles A, B, C and D is 240° which is less than 360°


(b) A + B + C + D = 240°.
And a Year 7 student is expected to comprehend this.
Anyway, thank you for your input, I appreciate it.
Re: A Question For The Mathematicians In The House by Timijo(m): 9:01am On Feb 21, 2022
dejt4u:
very simple question.

Split the shape into two triangles. If the angle at the middle is O.
Then we have triangles ADO and BCO.

But <C is 60° (i.e, (360 - (120×2))/2),
I have a question here.
How does the rule of the sum of angles at a point/reflex angles affect
angle C? It seems logical but I'm not convinced.
Never mind, I got the reason behind that now. It's even straightforward when the diagram is considered as a quadrilateral. Thanks again.

Therefore, A + D + 60° = 180°
A + D = 120°.......eqn 1
Similarly, B + C + 60° = 180°
B + C = 120° ........eqn 2

(a) sum of angles A, B, C and D can be established by adding eqn 1 and eqn 2 together. That is,
A + D + B + C = 120 + 120,
A + B + C + D = 240°.
therefore, the sum of angles A, B, C and D is 240° which is less than 360°


(b) A + B + C + D = 240°.

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