7. In A ABC, bisectors of A and B intersect at point O. If C = 70°. Find measure of AOB.
8. In Figure 3.11, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ZBPQ and ZPQD respectively. Prove that mZPTQ = 90°.
9. Using the information in figure 3.12, find the measures of a, b and c.
Answers
Answer:
Hey army here is your answer
7)ABC is a triangle. Bisectors of ∠A and ∠B intersect at point O. So, a measure of ∠C = 70°. Hence, the measure of ∠AOB = 125°.
8) Given: line AB || line CD and line PQ is the transversal.
ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively.
To prove: m ∠PTQ = 90°
Proof: ∠TPB = (1/2) ∠TPQ = ∠BPQ …(i) [Ray PT bisects ∠BPQ]
∠TQD = ∠TQP = (1/2) ∠PQD ….(ii)
[Ray QT bisects ∠PQD]
line AB || line CD and line PQ is their transversal. [Given]
∴∠BPQ + ∠PQD = 180° [Interior angles]
∴ (1/2) (∠BPQ) + (1/2) (∠PQD) = (1/2) x 180°
[Multiplying both sides by (1/2)]
∠TPQ + ∠TQP = 90°
In ∆PTQ,
∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 90° + ∠PTQ = 180° [From (iii)]
∴ ∠PTQ = 180° – 90° = 90°
∴ m ∠PTQ = 90°
Step-by-step explanation:
I hope this will helpful to you
Answer:
Sum of interior angles x & y is 180
o
on same side of line
x+y=180
o
…………(i)
In triangle PSR sum of angle is 180
o
2
x
+
2
y
+∠PSR=180
o
∠PSR=180−
2
1
(x+y)
From (i) x+y=180
∠PSR=90
o
.,
