In ΔPQR,
Solution:
[Pythagoras theorem]
PR2
=
PQ2
+
QR2
∴
(13)2
=
(12)2
+
QR2
∴
169
–
144
=
QR2
∴
=
5 cm
QR
EXERCISE 8.1
∠Q = 90°
Q
R
P
12 cm
13 cm
∴
=
25
QR2
5 cm
Q.2) In the given figure, find tan P – cot R.
cot R =
5
12
QR
PQ
tan P =
5
12
QR
PQ
∴
tan P =
∴
cot R =
=
tan P – cot R
5
12
5
12
–
∴
=
tan P – cot R
0