I’m revising circle theorems to strengthen my fundamentals, and I’m stuck on a small tangle of theorems.
Setup: I drew a circle with points A and B on the circumference so AB is a chord. A tangent touches the circle at A and meets the extension of AB at T (outside the circle). Point C is another point on the circle (not A or B), on the far side of chord AB. I’m given ∠TAB = 41°, and I’m meant to find ∠ACB (the angle at C subtended by chord AB).
I keep mixing up “angles in the same segment are equal,” the alternate segment theorem, and the cyclic quadrilateral 180° thing. My brain does a little dance and I lose track of which one actually connects the tangent-chord angle to the angle at C.
My (wrong) attempt: I treated the tangent as perpendicular to the chord (oops), so I set ∠TAB = 90°. Then I did 90° − 41° = 49°, and declared ∠ACB = 49° because I thought the angles “in the same segment” share what’s left. That’s clearly off, but I can’t seem to untangle it.
Question: Which specific circle theorem should I apply first to relate ∠TAB and ∠ACB in this configuration, and how should I state it so I pick the correct angle at C? A quick nudge on the right theorem and why my 49° logic fails would really help.
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3 Responses
Start with the Alternate Segment Theorem. It says: the angle between a tangent and a chord equals the angle in the opposite (alternate) segment subtended by that chord. Here the angle between the tangent AT and the chord AB is ∠TAB. Since C is on the far side of AB, ∠ACB is exactly the angle in the alternate segment subtended by AB. So ∠ACB = ∠TAB = 41°.
Why 49° is not right: a tangent is perpendicular to the radius at the point of contact, not to an arbitrary chord. So there is no 90° with AB unless AB is a diameter through A. Also, there are two inscribed angles subtended by the same chord AB-one on each side of AB. The one on the far side (your C) equals 41° by the Alternate Segment Theorem. The one on the near side is supplementary to it: 180° − 41° = 139°, which you can justify via the “angles in the same segment” idea or by opposite angles in a cyclic quadrilateral summing to 180°. Simple example: given ∠TAB = 41°, then ∠ACB (far side) = 41°, while for a point D on the near side, ∠ADB = 139°.
I’d start with the alternate segment theorem: the angle between a tangent and a chord at the point of contact equals the angle subtended by that chord in the opposite segment of the circle. Here, the tangent at A and the chord AB form ∠TAB, so the angle at any point C on the far side of AB that subtends AB (i.e., ∠ACB) must be equal to ∠TAB. Therefore ∠ACB = 41°. The 49° idea breaks because the tangent is perpendicular to the radius AO, not to the chord AB, so there is no right angle at A involving AB. Also, “angles in the same segment are equal” only compares two angles at the circumference that both subtend the same chord; you can’t use it directly with ∠TAB, since that angle isn’t at the circumference. Once you’ve used the alternate segment theorem to get one angle on the circumference (41°), the “same segment” fact would tell you any other angle on that same arc subtending AB is also 41°, but you don’t need the cyclic-quadrilateral 180° result here. Hope this helps!
Start with the alternate segment theorem (also called the tangent–chord theorem): the angle between a tangent and a chord at the point of contact equals the angle in the opposite segment of the circle subtended by that chord. In your picture, ∠TAB is the angle between the tangent AT and the chord AB at A, so it “mirrors” across the circle to the angle at any point C on the far side that subtends AB-i.e., ∠ACB. That means ∠ACB = ∠TAB = 41°. Your 49° detour came from treating the tangent as perpendicular to the chord; a tangent is perpendicular to the radius OA at A, not to an arbitrary chord AB (only if AB happened to be the diameter through A would that be 90°). A quick worked example: given ∠TAB = 41°, immediately write ∠ACB = 41° by the alternate segment theorem; as a check, pick any other point D on the same arc as C and you’ll also get ∠ADB = 41° by the “angles in the same segment are equal” theorem. The cyclic quadrilateral 180° rule is for four points all on the circle, so it’s not the first tool here. Nice visual walkthroughs: Corbettmaths on the Alternate Segment Theorem – https://corbettmaths.com/2013/05/30/alternate-segment-theorem/