Angle in a Semicircle

\( \angle\text{ in a semicircle }=90^{\circ} \)
Circle Theorems GCSE
Question 7 of 20
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\( AB\text{ is a diameter. The tangent at }A\text{ meets the extension of }BC\text{ at }T.\;\text{If }\angle BAC=32^{\circ},\text{ find }\angle ATB. \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Use the semicircle right angle, triangle sum to get \(\angle ABC\), AST for \(\angle TAB\), then sum angles in \(\triangle ATB\).

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.

  1. By the semicircle theorem, \(AB\) a diameter ⇒ \(\angle ACB=90^\circ\).
  2. Triangle sum in \(\triangle ABC\): \(\angle ABC = 180^\circ-90^\circ-27^\circ = 63^\circ\).
  3. 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 \(\angle TAB=90^\circ\).
  4. Because \(T\) lies on the extension of \(BC\), the interior angle at \(B\) of \(\triangle ATB\) is \(\angle ABT = \angle ABC = 63^\circ\).
  5. 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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  1. 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°
  2. 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°
  3. 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.
  4. 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°
  5. 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°
  6. 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: \( x=17 \)
  7. 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°
  8. 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°
  9. 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°
  10. 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: \( x=40 \)