Question 15868 – STQC-NIELIT SC-B 2021
December 7, 2023Operating-Systems
December 8, 2023Question 15872 – STQC-NIELIT SC-B 2021
In a parallelogram ABCD, AP and BP are the angle bisectors of ∟DAB znc ∟ABC. Find ∟APB:
Correct Answer: B
Question 22 Explanation:
we know that, ∟DAB+ ∟ ABC=180
dividing by 2,
½ ∟ DAB+½ ∟ ABC= 180/2
∟ PAB + ∟PBA=90. ………(1 say)
hence, in triangle APB,
∟PAB +∟PBA + ∟APB=180. (Angle sum prop)
from 1,
angAPB+90= 180
ang APB=90
dividing by 2,
½ ∟ DAB+½ ∟ ABC= 180/2
∟ PAB + ∟PBA=90. ………(1 say)
hence, in triangle APB,
∟PAB +∟PBA + ∟APB=180. (Angle sum prop)
from 1,
angAPB+90= 180
ang APB=90
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