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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? a32° b6° c24° d12°

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 th class=

Respuesta :

Step 1

One of the theorems of a circle states that the angles in the same segments or on the same chord are equal.

First, one theorem states that;

[tex]\frac{Arc(AB)-Arc(AX)}{2}=\hat{ACB}[/tex]

Step 2

[tex]\begin{gathered} Arc\text{ AB=152}^o \\ \frac{152-Arc(AX)}{2}=\hat{ACB} \end{gathered}[/tex]

Next, the Inscribed Angle Theorem states that;

[tex]\frac{Arc(AX)}{2}=\hat{ABX}[/tex][tex]Arc(AX)=2(64)=128^o[/tex]

Therefore the answer; Option D

[tex]\begin{gathered} \frac{152-128}{2}=\hat{ACB} \\ \hat{ACB}=12^o \end{gathered}[/tex]

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