he area of each figure using the formula,
10 m
m
A = (bl+ b2h
2
15 m
A = 3 (D x h)
Answers
Answer:
The value of \tan 60 ^ { \circ } = \sqrt { 3 }tan60
∘
=
3
To find:
The value of \tan 60 ^ { \circ }tan60
∘
Solution:
We know that
\tan ( A + B ) = \frac { \tan A + \tan B } { ( 1 - \tan A \cdot \tan B ) }tan(A+B)=
(1−tanA⋅tanB)
tanA+tanB
When A=B
\tan ( A + A ) = \frac { \tan A + \tan A } { 1 - \tan ^ { 2 } A }tan(A+A)=
1−tan
2
A
tanA+tanA
By using A = 30^{\circ}30
∘
\begin{gathered}\begin{array} { c } { \tan 60 ^ { \circ } = \frac { \tan 30 ^ { \circ } + \tan 30 ^ { \circ } } { 1 - \tan ^ { 2 } 30 ^ { \circ } } } \\\\ { \tan 60 ^ { \circ } = \frac { \frac { 1 } { \sqrt { 3 } + } \left( \frac { 1 } { \sqrt { 3 } } \right) } { \left( 1 - \left( \frac { 1 } { \sqrt { 3 } } \right) ^ { 2 } \right) } = \frac { \left( \frac { 2 } { \sqrt { 3 } } \right) } { \frac { 2 } { 3 } } = \frac { 2 } { \sqrt { 3 } } \times \frac { 3 } { 2 } = \sqrt { 3 } } \end{array}\end{gathered}
tan60
∘
=
1−tan
2
30
∘
tan30
∘
+tan30
∘
tan60
∘
=
(1−(
3
1
)
2
)
3
+
1
(
3
1
)
=
3
2
(
3
2
)
=
3
2
×
2
3
=
3
