Question

how do we relate tangent of a circle and tangent of an angle?

Answer 1

Relation between tangent of a circle and tangent of an angle:

Consider a unit circle on a cartesian plane. (A unit circle is one which has a radius of 1 unit.)

Now draw a right angled triangle making a central angle 'x' to the segment PB.

[Refer to the attachment 1]

Now since the circle is a unit circle, BO = 1.

$\implies \sin(x) = \sf \dfrac{Perpendicular}{Hypotenuse}$

$\implies \sin(x) = \sf \dfrac{AB}{BO}$

$\implies \sin(x) = \sf \dfrac{AB}{1}$

$\boxed{ \implies \sin(x) = \sf AB}$

Also,

$\implies \cos(x) =\sf \dfrac{Base}{Perpendicular}$

$\implies \cos(x) =\sf \dfrac{OA}{OB}$

$\implies \cos(x) =\sf \dfrac{OA}{1}$

$\boxed{ \implies \cos(x) =\sf OA}$

Now draw a tangent line QP, where point of tangency at P.

[Refer to the attachment 2]

Now try to find the similarity in ∆BAO and ∆QPO.

• ∠BOA is common
• ∠BAO = ∠QPO = 90°

Therefore both triangles are similar by (AA).

Note: Tangent of a circle makes 90° angle with the radius at the point of tangency.

Now since ∆BAO and ∆QPO are similar triangles, the ratio of pair of sides will be equal.

This implies that $\dfrac{BA}{AO} = \dfrac{QP}{OP}$

Further this implies that $\dfrac{\sin(x)}{\cos(x)} = \dfrac{QP}{1}$.

Therefore we arrive at a result:-

$\underline{ \boxed{\tan(x) = QP}}$

Conclusion:

Length of a line segment tangent to a circle that connects to a central angle is the tangent of the central angle.

This is the relation between tangent used in Geometry and in Trigonometry. And yes this is the origin of trigonometric tangent.

$\rule{290}{1}$

[Good to see this question on brainly, I used to have the same query. Can you also relate trigonometric secant (sec θ) with secant line on a circle?]