Question

AC and AD are tangents to a circle with center O at c and d respectively. if angle BCD=44, then find angle CAD,angle CBD,angle ACD.

Answer 1

Answer:

$\angle CAD=88^\circ$

$\angle CBD=92^\circ$

$\angle ACD=46^\circ$

Step-by-step explanation:

Let B is the center point of the circle.

From Point A, draw two tangents of a the circle AC and AD

Draw two lines BC and BD

Given that $\angle BCD=44^\circ$

In $\Delta BCD$ $BC=BD\text{ Both are radius of the circle}$

Therefore $\Delta BCD$ is a Isosceles Triangle, So

$\angle BCD=\angle BDC=44^\circ$

We know that

$\angle BCD+\angle BDC+\angle CBD=180^\circ\\\\\Rightarrow44^\circ+44^\circ+\angle CBD=180^\circ\\\\\Rightarrow\angle CBD=180^\circ-88^\circ=92^\circ$

$\therefore\angle CBD=92^\circ$

We know that radius is the perpendicular of the tangent

therefore, $\angle ADB =\angle ACB=90^\circ$

$\angle ACB=\angle ACD+\angle BCD\\\\\Rightarrow\angle ACD=\angle ACB-\angle BCD\\\\\Rightarrow\angle ACD=90^\circ-44^\circ=46^\circ$

$\therefore\angle ACD=46^\circ$

therefore, $\angle ADB =\angle ACB=90^\circ$

$\angle ADB=\angle ADC+\angle BDC\\\\\Rightarrow\angle ADC=\angle ADB-\angle BDC\\\\\Rightarrow\angle ADC=90^\circ-44^\circ=46^\circ$

$\therefore\angle ADC=46^\circ$

In $\Delta ACD$

$\angle CAD+\angle ACD+\angle ADC=180^\circ\\\\\Rightarrow\angle CAD=180^\circ-\angle ACD-\angle ADC\\\\\Rightarrow\angle CAD=180^\circ-46^\circ-46^\circ=88^\circ$

$\therefore\angle CAD=88^\circ$

Answer 2

Step-by-step explanation:

angle ACD equal 46 degree

angle CAD 88 degree

angle COD equal 92 degree

angle CBD equal 80 degree