Question

In the figure the radius of a circle with centre $C$ is $6cm$, line $AB$ is a tangent at $A$. Answer the following question:
(1) What is the measure of $\angle CAB$? Why?
(2) What is the distance of point $C$ from line $AB$? Why?
(3) $d(A,B)=6cm$, find $d(B,C)$
(4) What is the measure of $\angle ABC$? Why?

Answer 1

Join $A,C$ and $B,C$

(1) We know that, Tangent is always perpendicular to the radius at point of contact.

$\therefore \angle CAB={90}^{\circ }$

( 2 ) The distance of point $C$ from the line $AB$ is equal to the radius of the circle. ie., line $AC$

$AC=$ radius of circle $=6cm$

( 3 ) $AB=6cm$ and $AC=6cm$

$ABC$ is a right angles triangle.

By Pythagoras theorem,

$B{C}^2=A{B}^2+A{C}^2$

$B{C}^2=6^2+6^2=36+36$

$BC=\sqrt{72}=6\sqrt{2}cm$

$\therefore d(B,C)=6\sqrt{2}cm$

( 4 ) $ABC$ is an isosceles triangle.

$AB=AC$ [$\because AB=AC=6cm$]

$⟹\angle ABC=\angle ACB$ ------ Angles opposite to equal sides in a triangle are equal

In $△ABC$,

$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$

${90}^{\circ }+2\angle ABC={180}^{\circ }$

$2\angle ABC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

$\therefore \angle ABC=\frac{{90}^{\circ }}{2}={45}^{\circ }$