How to solve this question no. 10 of chapter 8. Exercise 8.1. plzzz solve it..
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PR + QR =25
PR=25 - QR................(1)
5^2 + QR^2 = PR^2..........(PYTHAGOREAN THEO)
THEN SUBSTITUTE PR,
=5^2 + QR^2
=(25-QR)^2
SOLVE FOR QR,
5^2+QR^2=625-50 QR + QR^2
5^2=625-50 QR
-600=-50 QR
QR=12
.°. PR=13
sin=12/13, cos=5/13, tan=12/5
hope it help u..........
PR=25 - QR................(1)
5^2 + QR^2 = PR^2..........(PYTHAGOREAN THEO)
THEN SUBSTITUTE PR,
=5^2 + QR^2
=(25-QR)^2
SOLVE FOR QR,
5^2+QR^2=625-50 QR + QR^2
5^2=625-50 QR
-600=-50 QR
QR=12
.°. PR=13
sin=12/13, cos=5/13, tan=12/5
hope it help u..........
riteshkumar92:
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Answered by
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