In a right angle triangle ∆ pqr , pq + qr = 49 cm and pr = 41 cm. Find the value of Cos^2 r - sin^2 - r ( given that q = 90° )
Answers
Answer:
R.E.F. Image.
Given ΔPQR is right angle triangle with QR=9cm
& PR−PQ=1cm
then PR=(1+PQ)
squaring both sides
(PR)
2
=(1+PQ)
2
=1+(PQ)
2
+2PQ
(PR)
2
−(PQ)
2
=1+2PQ
(QR)
2
=1+2PQ [PR
2
=PQ
2
+QR
2
]
81−1=2PQ
[PQ=40cm]
now, using Pythagoras theorem. or equation (1) we have
PR−PQ=1
[PR=41cm]
now sinR=
PR
PQ
=
41
40
cosR=
41
9
=
PR
QR
then sinR+cosR=
41
40
+
41
9
=
41
49
.
sinR+cosR=
41
49
.
Explanation:
hope it helps
Answer:
Given ΔPQR is right angle triangle with QR=9cm
& PR−PQ=1cm
then PR=(1+PQ)
squaring both sides
(PR)
2
=(1+PQ)
2
=1+(PQ)
2
+2PQ
(PR)
2
−(PQ)
2
=1+2PQ
(QR)
2
=1+2PQ [PR
2
=PQ
2
+QR
2
]
81−1=2PQ
[PQ=40cm]
now, using Pythagoras theorem. or equation (1) we have
PR−PQ=1
[PR=41cm]
now sinR=
PR
PQ
=
41
40
cosR=
41
9
=
PR
QR
then sinR+cosR=
41
40
+
41
9
=
41
49
.
sinR+cosR=
41
49
.
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