(07.01 LC) 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: The figure shows a circle with points A and B on it and point C outside it. Side BC of triangle ABC intersects the circle at point X. A tangent to the circle at point A is drawn from point C. Arc AB measures 152 degrees, and angle CBA measures 64 degrees. What is the measure of angle ACB?

Option (D)

Step-by-step explanation:

Measure of arc AB = 152°

"Since measure of inscribed angle is half the measure of intercepted arc"

Measure of arc AX = 2(m∠ABX)

                               = 2 × 64°

                               = 128°

"When a tangent and secant intersect outside a circle, measure of angle between them will be half of the difference of the measures of intercepted arcs."

m∠ACB =

             =

             = 12°

Therefore, measure of angle ACB = 12° is the answer.

Option (D) is the correct option.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)