Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ∠QPR respectively. Ray PB and ray PA are perpendicular to each other. Draw a figure showing all these rays and write - (i) A pair of complementary angles (ii) A pair of supplementary angles. (iii) A pair of congruent angles.
Answers
Given:
Ray PQ and PR are perpendicular each other. Point B is the interior and Point A is the exterior of ∠QPR.
Ray PB and PA are perpendicular to each other.
Now, we shall find
(i) A pair of Complementary angles:
(a) ∠QPB and ∠BPR because their sum is 90°
(b) ∠BPR and ∠RPA because their sum is 90°
(ii) A pair of Supplementary angles:
∠QPR and ∠BPA because their sum is 180°
(iii) A pair of congruent angles:
(1) ∠BPA ≡ ∠QPR -------- because each is 90°
(2) ∠BPA = ∠ QPR --------- as given
so, ∠RPA + ∠BPR = ∠BPR + ∠QPB------- (using angle addition property)
so, ∠ RPA = ∠QPB
Hence, ∠ RPA ≡ ∠QPB
Answer
Ray PQ and PRare perpendicular each other. Point B is the interior and Point A is the exterior of ∠QPR.
Ray PB and PA are perpendicular to each other.
(i) A pair of Complementary angles:
⇒ ∠QPB and ∠BPR because their sum is 90
o
⇒ ∠BPR and ∠RPA because their sum is 90
o
(ii) A pair of Supplementary angles:
⇒ ∠QPR and ∠BPA because their sum is 180
o
(iii) A pair of congruent angles:
⇒ ∠BPA≡∠QPR [ Each angle is 90
o
⇒ ∠BPA=∠QPR
So, ∠RPA+∠BPR=∠BPR+∠QPB [ Using angle addition property ]
So, ∠RPA=∠QPB
Hence, ∠RPA≡∠QPB
