Question

The figure below shows a triangle with vertices A and B on a circle and vertex C outside it. Side AC is tangent to the circle. Side BC is a secant intersecting the circle at point X. What is the measure of angle ACB? 32° 60° 28° 16°

Answer 1

Can someone help please

Answer 2

Okay, so
BX = BA (Radii of the circle)
We know that because it is a triangle, the sum of angles MUST be 180 degrees.
Also, we know that the radii of a circle are always perpendicular to a tangent drawn at that point.
Let's name the center as D.
So, we now have a new triangle ADB, where
mDBA = mDAB = x (let's assume it as x, we know both angles are equal because side BD and DA are equal as they are radii of the circle).
In triangle DBA,
mBDA = 176 degrees (given), and
mDBA = mDAB = x
So, by angle sum property,
x+x+176 = 180
2x+176 = 180
2x=180-176
2x=4
x=4/2 = 2 degrees.
Therefore angle mBAC = 90+2 = 92 degrees.

Now apply the angle sum property:
mCBA + mBAC + mACB = 180
56 + 92 + x = 180
148 + x = 180
x = 180 - 148
x = 32 degrees.

Therefore, measure of ACB is 32 degrees (first choice).