Question

in the circl below, AD is diameter and AB is tangent at A. Suppose mADC=228*, find the measures of mCAB and mCAD. Type your numerical answers without units in each blank.

Answer 1

Given:

AD is diameter of the circle, AB is the tangent, and measure of arc ADC is 228 degrees.

To find:

The $m\angle CAB$ and $m\angle CAD$.

Solution:

AD is diameter of the circle. So, the measure of arc AD is 180 degrees.

$m(arcADC)=m(arcAD)+m(arcDC)$

$228^\circ=180^\circ+m(arcDC)$

$228^\circ-180^\circ+=m(arcDC)$

$48^\circ+=m(arcDC)$

The measure inscribed angle is half of the corresponding subtended arc.

$m\angle CAD=\dfrac{1}{2}\times m(arcDC)$

$m\angle CAD=\dfrac{1}{2}\times 48^\circ$

$m\angle CAD=24^\circ$

AB is the tangent. So, $m\angle BAD=90^\circ$ because radius is perpendicular on the tangent and the point of tangency.

$m\angle BAD=m\angle CAB+m\angle CAD$

$90^\circ=m\angle CAB+24^\circ$

$90^\circ -24^\circ=m\angle CAB$

$66^\circ=m\angle CAB$

Therefore, $m\angle CAB=66^\circ$ and $m\angle CAD=24^\circ$.