Question

in the given figure,AB is the tangent to the circle with centre O such that OB=6cm.then,AB is equal to​

Answer 1

$\large\underline{\sf{Given- }}$

A circle with centre O,

such that

• AB is the tangent

• OB=6cm

• ∠ ABO = 30°

$\large\underline{\sf{To\:Find - }}$

• Length of tangent, AB.

$\large\underline{\sf{Solution-}}$

Since,

• AB is tangent to a circle and OA Is radius,

$\rm :\implies\:AO \: \perp \: AB$

$\rm :\implies\: \triangle \: ABO \: is \: right \: angle \:$

Now,

$\rm :\longmapsto\:In \:right \: \triangle \: ABO$

$\rm :\longmapsto\:cos \: \angle \: ABO = \dfrac{AB}{BO}$

$\rm :\longmapsto\:cos \: 30 \degree \: = \dfrac{AB}{6}$

$\rm :\longmapsto\:\dfrac{ \sqrt{3} }{2} = \dfrac{AB}{6}$

$\bf\implies \:AB \: = \: 3 \sqrt{3} \: cm$

Additional Information :-

1. The length of tangents from an external point to a circle are equal.

2. It is perpendicular to the radius of the circle at the point of tangency.

3. The tangent touches the circle at only one point.

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$