EXERCISE 8.1
Q.9) In triangle ABC, right angled at B, if tan A =
1
i) sin A cos C + cos A sin C
B
C
A
Solution:
1
tan A =
…(i)
BC
AB
∴
1
=
[From (i) and (ii)]
BC
AB
tan A =
…(ii)
=
AB
=
k
BC
[Pythagoras theorem]
Let the non zero common multiple be k.
AC2
=
AB2
+
BC2
∴
AC2
=
+
(1k)2
∴
AC2
=
3k2
+
k2
∴
AC2
=
4k2
∴
AC
=
2k
2k
k
In ΔABC, ∠B = 90º
EXERCISE 8.1
Q.9) In triangle ABC, right angled at B, if tan A =
1
i) sin A cos C + cos A sin C
Solution:
B
C
A
2k
k
sin A =
BC
AC
1k
2k
∴
sin A =
1
2
∴
sin A =
cos C =
BC
AC
1k
2k
∴
cos C =
1
2
∴
cos C =
cos A =
AB
AC
2k
∴
cos A =
2
∴
cos A =
sin C =
AB
AC
2k
∴
sin C =
2
∴
sin C =
EXERCISE 8.1
Q.9) In triangle ABC, right angled at B, if tan A =
1
i) sin A cos C + cos A sin C
Solution:
B
C
A
2k
k
sin A cos C + cos A sin C
=
1
2
×
1
2
+
2
×
2
=
1
4
+
3
4
=
1
+
3
4
=
4
4
∴
sin A cos C + cos A sin C
=
1
1
2
sin A =
1
2
cos C =
2
cos A =
2
sin C =
EXERCISE 8.1
Q.9) In triangle ABC, right angled at B, if tan A =
1
ii) cos A cos C – sin A sin C
Solution:
B
C
A
2k
k
cos A cos C – sin A sin C
=
2
×
1
2
–
1
2
×
2
=
4
–
4
=
0
∴
sin A cos C + cos A sin C
=
0
1
2
sin A =
1
2
cos C =
2
cos A =
2
sin C =