Question
(a) Derive the angle sum property of the triangle using the exterior angle property.
(b) AE is the angular bisector. ∠4=140∘,∠2=60∘,∠EAC=x. Find x.
[4 MARKS]
(b) AE is the angular bisector. ∠4=140∘,∠2=60∘,∠EAC=x. Find x.
[4 MARKS]
Answer:
:
(a) Steps: 1 Mark
Proof: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a)
∠1, ∠2 and ∠3 are angles of triangle ABCand 4 is the exterior angle when BC is extended to D.
∠1 + ∠2 = ∠4 (Exterior Angle Property)
∠1 + ∠2 + ∠3 = ∠4 + ∠3 (Adding ∠3 to both sides)
Also, ∠3 + ∠4 = 180o (Linear Pair of angles)
∴ ∠1 + ∠2 + ∠3 = 180o
(b)
∠1 + ∠2 = ∠4 (Exterior Angle Property)
∠1+60∘=140∘
∠1=140∘−60∘=80∘
Since AE is the angular bisector therefore ∠EAC=∠12
⇒x=802=40∘
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:
(a) Steps: 1 Mark
Proof: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a)
∠1, ∠2 and ∠3 are angles of triangle ABCand 4 is the exterior angle when BC is extended to D.
∠1 + ∠2 = ∠4 (Exterior Angle Property)
∠1 + ∠2 + ∠3 = ∠4 + ∠3 (Adding ∠3 to both sides)
Also, ∠3 + ∠4 = 180o (Linear Pair of angles)
∴ ∠1 + ∠2 + ∠3 = 180o
(b)
∠1 + ∠2 = ∠4 (Exterior Angle Property)
∠1+60∘=140∘
∠1=140∘−60∘=80∘
Since AE is the angular bisector therefore ∠EAC=∠12
⇒x=802=40∘
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