Answer:
ANgle CAO measure [tex]21^o[/tex]
Step-by-step explanation:
The important procedure here is to work with the three "isosceles" triangles that are formed inside the circle all of them having a common vertex at the center of the circle (point O). Notice that all these three triangles must have two sides equal (therefore be isosceles triangles) because two sides measure exactly the radius of the circle. Have in mind then that these equal sides must oppose equal angles.
1) Triangle COB has two angles which measure [tex]32^o[/tex], due to the fact that they oppose the sides equal to the radius. Then, the third angle of the triangle (the one at the center O must measure:
[tex]180^o-32^o-32^o=116^o[/tex]
2) For triangle AOB, we can find the value at the acute angle [tex]\angle \,A[/tex] (with vertex at point A), considering that it can be obtained by subtracting [tex]53^o[/tex] from [tex]90^o[/tex], since the radius of the circle is always perpendicular to the tangent line at the point where the radius meets the tangent. Then this acute angle must measure:
[tex]90^o-32^o=37^o[/tex]
Now, the other angle opposite to the radius must also measure [tex]37^o[/tex], so we can find the measure of the larger angle in the triangle (that with vertex at point O) by subtracting these two small angles from [tex]180^o[/tex]:
[tex]180^o=37^o-37^o=106^o[/tex]
3) Finally we get to study the third triangle (COA):
We now know the value of the central angle (that with vertex at point O), since it should be equal to the full circle [tex](360^o)[/tex] minus the other two central angles we calculated above:
[tex]360^o-116^o-106^o=138^o[/tex]
so, now we can determine the measure of the addition of the other two small (and equal) angles of the triangle:
[tex]180^o-138^o=42^o[/tex]
Therefore, each of the angles should measure half of this value. that is each of the small acute angles measures [tex]21^o[/tex]
This is therefore the measure of the requested angle CAO.
