- Given Information:
- ππ=2PQ=2 cm
- ππ
=1QR=1 cm
- ππ
PR is perpendicular to ππQS
- ππ
=π
πPR=RS
- Calculate ππ
PR:
ππ
=π
π=π₯ (since ππ
=π
π)PR=RS=x (since PR=RS)
Using Pythagorean theorem in β³πππ
β³PQR:
ππ2=ππ
2+ππ
222=π₯2+124=π₯2+1π₯2=3π₯=3PQ2=PR2+QR222=x2+124=x2+1x2=3x=3β
- Determine β πππβ QPS: Since ππ
=π
πPR=RS and β³ππ
πβ³PRS is an isosceles right triangle:
β ππ
π=90β (because itβs isosceles and right)β PRS=90β (because itβs isosceles and right)
Since ππ
PR is perpendicular to ππQS, in β³πππβ³PQS, we have:
β πππ=β ππ
π+β ππ
π=90β+45β=135ββ QPS=β PRQ+β PRS=90β+45β=135β
Thus:
β πππ=180ββ135β=45ββ QPS=180ββ135β=45β
But re-evaluating, from isosceles setup splitting:
Angle sum and misread reassessment simplifies correct final:
180ββ90ββ75β180ββ90ββ75β
Correct angle β πππ=75ββ QPS=75β.
The answer is:
D. 75Β°