Question - Calculating the Angle of a Tangent-Secant Triangle in a Circle Geometry Problem
Solution:
Let \( \angle BAX = y \) and \( \angle BFX = z \).
Since AE is tangent to the circle at B, \( \angle AEB = 90^\circ \).
By the alternate segment theorem, \( \angle ABX = \angle AEB = 90^\circ \).
Consider triangle ABX: \( x + y + 90^\circ = 180^\circ \)
\( x + y = 90^\circ \) ...(1)
Since AE and AX are tangent to the circle at E and X, \( \angle EAX = 80^\circ \).
In triangle AEX: \( y + 80^\circ + 90^\circ = 180^\circ \)
\( y = 180^\circ - 80^\circ - 90^\circ \)
\( y = 10^\circ \) ...(2)
Using (1) and (2), we find \( x \):
\( x + 10^\circ = 90^\circ \)
\( x = 90^\circ - 10^\circ \)
\( x = 80^\circ \) ...(3)
Since BF is tangent to the circle at F and BC is a secant, \( \angle BFX = \angle BCX \) (angles in the alternate segment).
Triangle BCF is isosceles (BF = BC as radii of the same circle), so \( \angle BCF = \angle BFC \).
Consider the sum of angles in triangle BCF: \( z + z + x = 180^\circ \)
\( 2z = 180^\circ - x \)
Using the value of \( x \) from (3): \( 2z = 180^\circ - 80^\circ \)
\( 2z = 100^\circ \)
\( z = 50^\circ \)
Finally, \( \angle AFX = x - z \)
\( \angle AFX = 80^\circ - 50^\circ \)
\( \angle AFX = 30^\circ \)
Tutor Solution:
First, let's establish a few properties related to tangents and circles: - The tangent to a circle is perpendicular to the radius drawn to the point of tangency. - If two tangents are drawn to a circle from an external point, they are equal in length.
From the information given, we have that AE and AX are tangents to the two circles from point A. Therefore, AE = AX by the property that two tangents drawn from an external point to a circle are equal.
∠EAX is 80°, as given. This angle is an external angle to triangle AEX, where EX is a line segment that joins the points of tangency of AE and AX. Therefore, ∠EAX = ∠AEX + ∠AXE (because an exterior angle of a triangle is equal to the sum of the opposite interior angles).
Since AE = AX, triangle AEX is an isosceles triangle with AE = AX. This implies that ∠AEX = ∠AXE. So we can set up the equation: ∠EAX = ∠AEX + ∠AXE 80° = y + y 80° = 2y y = 40°
Now let's consider quadrilateral ABFX. All four points lie on the outer circle, so this quadrilateral is cyclic. In a cyclic quadrilateral (a four-sided figure where all the vertices lie on a single circle), opposite angles sum up to 180°.
Since y (40°) is the measure of angle AXB (because AX is tangent to the inner circle at point X and BX is the radius to the point of tangency), the opposite angle AFB must sum with 40° to make 180°.
Therefore, we have: ∠AXB + ∠AFB = 180° 40° + ∠AFB = 180° ∠AFB = 180° - 40° ∠AFB = 140°
We also have that FB is tangent at point F, and AB is the radius to that point of tangency, which means that ∠AFB is a right angle and hence, AFB is a 90° angle. Now, we can use the information that the sum of angles in triangle AFB equals 180° to find the remaining angle at AFX:
∠AFX + ∠AFB + ∠BAX = 180° ∠AFX + 90° + 40° = 180° ∠AFX + 130° = 180° ∠AFX = 180° - 130° ∠AFX = 50°
Therefore, the measure of angle AFX is 50°.
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