Answer:
[tex]\sqrt3[/tex]
Step-by-step explanation:
To find:
The value of [tex]tan60^\circ[/tex] = ?
Solution:
Kindly consider the equilateral [tex]\triangle ABC[/tex] as attached in the answer area.
Let the side of triangle = [tex]a[/tex] units
Let us draw the perpendicular from vertex A to side BC.
It will divide the side BC in two equal parts.
i.e. BD = DC = [tex]\frac{a}{2}[/tex]
Using Pythagorean Theorem in [tex]\triangle ABD[/tex]:
[tex]\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}[/tex]
Side AD = [tex]\frac{\sqrt3}{2}a[/tex]
Using Trigonometric ratio:
[tex]tan\theta = \dfrac{Perpendicular}{Base}[/tex]
[tex]tanB = \dfrac{AD}{BD}[/tex]
Putting the values of AD and BD:
[tex]tan60^\circ=\dfrac{\frac{\sqrt3}{2}a}{\frac{1}{2}a}\\\Rightarrow tan60^\circ = \bold{\sqrt3}[/tex]
