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Problems
If
1
+
sin
6
∘
cos
6
∘
=
tan
A
=
1
+
sin
B
1
−
sin
B
\frac{1 + \sin 6^\circ}{\cos 6^\circ} = \tan A = \sqrt{\frac{1 + \sin B}{1 - \sin B}}
c
o
s
6
∘
1
+
s
i
n
6
∘
=
tan
A
=
1
−
s
i
n
B
1
+
s
i
n
B
…
Problem
If
1
+
sin
6
∘
cos
6
∘
=
tan
A
=
1
+
sin
B
1
−
sin
B
\frac{1 + \sin 6^\circ}{\cos 6^\circ} = \tan A = \sqrt{\frac{1 + \sin B}{1 - \sin B}}
c
o
s
6
∘
1
+
s
i
n
6
∘
=
tan
A
=
1
−
s
i
n
B
1
+
s
i
n
B
; where
A
A
A
and
B
∈
(
0
,
90
∘
)
B \in (0, 90^\circ)
B
∈
(
0
,
9
0
∘
)
, then a.
A
=
8
B
A = 8B
A
=
8
B
b.
8
A
=
B
8A = B
8
A
=
B
c.
A
−
7
B
=
6
∘
A - 7B = 6^\circ
A
−
7
B
=
6
∘
d.
A
+
B
=
54
∘
A + B = 54^\circ
A
+
B
=
5
4
∘
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