Angle in a Semicircle – Formula Practice

\( AB\text{ is a diameter and }C\text{ lies on the circle. If }\angle ABC=62^{\circ},\text{ find }\angle BAC. \)

Explanation

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Extended example

Problem. \(AB\) is a diameter of a circle with centre \(O\). A point \(C\) lies on the circle above the diameter. The tangent at \(A\) meets the extension of \(BC\) at \(T\). If \(\angle BAC=27^\circ\), find \(\angle ATB\).

Solution.

By the semicircle theorem, \(AB\) a diameter ⇒ \(\angle ACB=90^\circ\).
Triangle sum in \(\triangle ABC\): \(\angle ABC = 180^\circ-90^\circ-27^\circ = 63^\circ\).
Alternate Segment Theorem for chord \(AB\): the angle between the tangent at \(A\) and \(AB\) equals the angle in the alternate segment that subtends \(AB\). Since \(\angle ACB=90^\circ\), we get \(\angle TAB=90^\circ\).
Because \(T\) lies on the extension of \(BC\), the interior angle at \(B\) of \(\triangle ATB\) is \(\angle ABT = \angle ABC = 63^\circ\).
Now use the triangle angle sum in \(\triangle ATB\): \[ \angle ATB = 180^\circ - \angle TAB - \angle ABT = 180^\circ - 90^\circ - 63^\circ = 27^\circ. \]

Answer: \(\angle ATB = 27^\circ\).

Worked examples

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AB is a diameter of a circle and C is a point on the circle. If \(\angle BAC=31^{\circ}\), find \(\angle ABC\).
1. Angle in a semicircle: since AB is a diameter, \(\angle ACB=90^{\circ}\).
2. Triangle sum: \(\angle ABC = 180^{\circ}-90^{\circ}-31^{\circ}=59^{\circ}\).
Answer: 59°
AB is a diameter and C lies on the circle. Find \(\angle ACB\).
1. Angle in a semicircle: the angle subtending a diameter at the circumference is a right angle.
Answer: 90°
In a circle, \(\angle ACB=90^{\circ}\). What can you conclude about \(AB\)?
1. Converse of the semicircle theorem: a right angle at the circumference implies the opposite side is a diameter.
Answer: AB is a diameter.
AB is a diameter. The tangent at A meets the extension of BC at T. If \(\angle BAC=27^{\circ}\), find \(\angle ATB\).
1. Semicircle: \(\angle ACB=90^{\circ}\).
2. Triangle \(ABC\): \(\angle ABC=180-90-27=63^{\circ}\).
3. Alternate Segment Theorem with chord AB: \(\angle TAB=\angle ACB=90^{\circ}\).
4. Since T lies on the extension of BC, the interior angle at B of triangle ATB is \(\angle ABT=\angle ABC=63^{\circ}\).
5. Triangle sum in \(\triangle ATB\): \(\angle ATB=180-90-63=27^{\circ}\).
Answer: 27°
AB is a diameter. If \(\angle ABC=52^{\circ}\), find \(\angle BAC\).
1. Angle in a semicircle: \(\angle ACB=90^{\circ}\).
2. Triangle sum: \(\angle BAC=180-90-52=38^{\circ}\).
Answer: 38°
AB is a diameter. If \(\angle ABC=(3x+8)^{\circ}\) and \(\angle BAC=(2x-3)^{\circ}\), find \(x\).
1. In \(\triangle ABC\), \(\angle ACB=90^{\circ}\).
2. So the two acute angles sum to 90°: \((3x+8)+(2x-3)=90\).
3. Solve: \(5x+5=90\Rightarrow x=17\).
Answer: \( x=17 \)
Points C and D both lie on the same semicircle over diameter AB. Find \(\angle ADB\).
1. Any angle subtending a diameter is a right angle.
2. Therefore \(\angle ADB=90^{\circ}\) (and also \(\angle ACB=90^{\circ}\)).
Answer: 90°
AB is a diameter. What is the angle between the tangent at A and the chord AB?
1. Alternate Segment Theorem with chord AB: tangent–chord angle equals \(\angle ACB\) in the opposite segment.
2. Since AB is a diameter, \(\angle ACB=90^{\circ}\).
3. Hence the tangent is perpendicular to AB.
Answer: 90°
AB is a diameter. If \(\angle BAC=40^{\circ}\) and D is on the same semicircle as C, find \(\angle ABD\).
1. First \(\angle ACB=90^{\circ}\), so \(\angle ABC=180-90-40=50^{\circ}\).
2. Angles in the same segment: \(\angle ABD=\angle ACD=50^{\circ}\) since both subtend arc AD (with D chosen on the same semicircle).
Answer: 50°
AB is a diameter. Suppose \(\angle BAC=x^{\circ}\) and \(\angle ABC=(x+10)^{\circ}\). Find \(x\).
1. Semicircle: \(\angle ACB=90^{\circ}\).
2. Acute angles sum to 90°: \(x + (x+10) = 90\).
3. Solve: \(2x+10=90\Rightarrow x=40\).
Answer: \( x=40 \)