In an acute angled ΔABC, if tan (A + B – C) = 1,
sec (B + C – A) = 2, then find the value of C.
tan (A + B – C) = 1
∴
sec (B + C – A) = 2
sec
∴
Adding (i) and (ii), we get
tan
=
1
A
+
B
C
–
...(i)
=
45
=
2
B
+
C
A
–
...(ii)
=
60
Solution:
But,
But,
45º
?
60º
?
2B
=
105
∴
B
=
52.5
tan
(A
+
B
C)
–
=
tan
45º
A + B – C
B + C – A
�=
�45 +
�60
� +
Example:
...(iii)
In an acute angled ΔABC, if tan (A + B – C) = 1,
sec (B + C – A) = 2, then find the value of C.
Solution:
Adding (ii) and (iv), we get
52.5
∴
C
∴
C
∴
=
67.5
120
−
52.5
=
+
C
=
120
2B
+
2C
=
240
...(iv)
A
+
B
C
+
=
180
∴
B
+
C
=
120
B + C – A = 60 …(ii)
B = 52.5
(Angle sum property of a triangle)
B + C – A
A + B + C
=
�60 +
�180
� +
∴
(Dividing throughout by 2)
Example:
...(iii)
...[From (iii)