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primary 5 | Maths
| Angles & Geometry
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Could someone help and explain this question to me? Thank you
Date Posted: 3 years ago
Views: 655
Because AB = BC = CD, we can say that
1. Triangle ABC is isosceles, with angle CAB = angle ACB.
2. Triangle BCD is isosceles, with angle DBC = angle BDC.
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From point 1, we have angle ACB = 40°. Now,, if we notice, lines BED and AEC intersect at E. Have you learnt "vertically opposite angles"?
If we notice, angles AED and BEC are directly opposite one another, so we can say that angle BEC = 110°.
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Let's focus on triangle BEC now. Angle ECB is 40°. Angle BEC is 110°. That leaves...
Angle EBC = 180° - 40° - 110° = 30°.
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But from point 2, we have angle BDC = 30°.
So, focusing on triangle BCD, the marked angle
= 180° - 30° - 30° - 40°
= 80°
1. Triangle ABC is isosceles, with angle CAB = angle ACB.
2. Triangle BCD is isosceles, with angle DBC = angle BDC.
--------------------------------------------------------
From point 1, we have angle ACB = 40°. Now,, if we notice, lines BED and AEC intersect at E. Have you learnt "vertically opposite angles"?
If we notice, angles AED and BEC are directly opposite one another, so we can say that angle BEC = 110°.
--------------------------------------------------------
Let's focus on triangle BEC now. Angle ECB is 40°. Angle BEC is 110°. That leaves...
Angle EBC = 180° - 40° - 110° = 30°.
--------------------------------------------------------
But from point 2, we have angle BDC = 30°.
So, focusing on triangle BCD, the marked angle
= 180° - 30° - 30° - 40°
= 80°
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