Question

A B and C are points on a circle center O. TA is a tangent to the circle at A and OBT is a straight line. AC is a diameter and angle OTA=24 degrees. Calculate angle AOT angle ACB angle ABT​

Answer 1

Given:

$\text { TA is a tangent to the cirde at } A \text {. }$

$\therefore \text { angle } O A T=90^{\circ}$

$\therefore \text { angle } A O T=180^{\circ}-0 A T-{T}=180^{0}-90^{\circ}-24^{\circ}$

$=90^{\circ}-24^{\circ}=66^{\circ}$

$\because O A=O B \quad \therefore O \hat{A} B=OBA$

$\therefore \hat{O A} B=0 \hat{B A}=\left(180^{\circ}-A \hat{O} T\right) \div 2=\left(180^{\circ}-66^{\circ} \div 2\right.$

$=114^{\circ} \div 2=57^{\circ}$

$\because \text { A. B and C are points on a circle, c entre o. }$

$\begin{array}{l}\therefore A \hat{B C -}90^{\circ}\\\therefore A C B=180^{\circ}-A \hat{B C}=O\hat{A} B=180^{}-90^{\circ}-57^{\circ}\\=90^{\circ}-57^{0}=33^{\circ}\end{array}$

$ABT=180^{\circ}-0 \hat{B} A=180^{0}-57^{\circ}=123^{\circ}$

The angles are as follows:

angle AOT=66°

angle ACB=33°

angle ABT​=123°