Photo AI
8.2 In die diagram hieronder is RT 'n raaklyn aan die sirkel ADBC by punt C sodanig dat ∠TCB = 37° AC || DB 8.2.1 Skryf neer, met redes, VIER ander hoeke gelyk aan 37° - NSC Technical Mathematics - Question 8 - 2023 - Paper 2
Question 8
8.2 In die diagram hieronder is RT 'n raaklyn aan die sirkel ADBC by punt C sodanig dat ∠TCB = 37° AC || DB 8.2.1 Skryf neer, met redes, VIER ander hoeke gelyk aan ... show full transcript
Worked Solution & Example Answer:8.2 In die diagram hieronder is RT 'n raaklyn aan die sirkel ADBC by punt C sodanig dat ∠TCB = 37° AC || DB 8.2.1 Skryf neer, met redes, VIER ander hoeke gelyk aan 37° - NSC Technical Mathematics - Question 8 - 2023 - Paper 2
Step 1
8.2.1 Skryf neer, met redes, VIER ander hoeke gelyk aan 37°.
Answer
According to the properties of angles formed by parallel lines and a transversal, we can state the following:
- ∠TCB = 37° (given)
- By Alternate Angles, ∠ACB = 37°, because AC || DB.
- By Corresponding Angles, ∠TCA = 37°, as RC is a transversal.
- Also, ∠EDB = 37°, since DE is parallel to AC and DB acts as a transversal. Hence, we have established four angles equal to 37°.
Step 2
8.2.2 Toon Vervolgens dat ∠AEC || ∠BED.
Answer
To prove that ∠AEC is parallel to ∠BED, we can refer to the Converse of the Alternate Interior Angles Theorem. Since we've established that:
- AC || DB
- ∠ACB and ∠TCB are alternate interior angles. Thus, since ∠ACB is equal to ∠TCB, it follows that ∠AEC is parallel to ∠BED.
Step 3
8.2.3 Voltooi vervolgens die bewering AE × ED = ... × ...
Answer
Following the properties of similar triangles or areas, if the triangles or figures formed by the segments AE and ED maintain proportionality with other corresponding segments, it can be concluded that: AE × ED = AC × DB This completes the assertion by establishing a relationship between the segments due to the parallel lines.