Angle in a Semicircle

Worked example

1. 1

   $\angle CAB=37^{\circ }$ — The question. AB is a diameter, C is on the circle, and the angle at A is given.

2. 2

   $\angle ACB=90^{\circ }$ — C sits on the circle and looks at the diameter, so angle in a semicircle is 90°. Mark it on the diagram.

3. 3

   $\angle ABC=180^{\circ }-90^{\circ }-37^{\circ }$ — Angles in a triangle add up to 180°, so take away the two angles you know.

4. 4

   $\angle ABC=53^{\circ }$ — 180 − 90 = 90, then 90 − 37 = 53. Check: 37 + 90 + 53 = 180.

Worked example

1. 1

   $\angle BAO=58^{\circ }$ — The question. AC is a diameter, O is the centre, B is on the circle and OB is drawn. Find angle OBA and angle OCB.

2. 2

   $OA=OB$ — Both are radii of the same circle, so triangle OAB is isosceles.

3. 3

   $\angle OBA=58^{\circ }$ — In an isosceles triangle the two base angles are equal.

4. 4

   $\angle ABC=90^{\circ }$ — Angle in a semicircle is 90°, because AC is a diameter.

5. 5

   $\angle OBC=90^{\circ }-58^{\circ }=32^{\circ }$ — Angle ABC is split by OB into two parts, so take the 58° away from the 90°.

6. 6

   $OB=OC,\text{so}\angle OCB=32^{\circ }$ — Radii again, so triangle OBC is isosceles too. Its base angles are equal.

Worked example

1. 1

   $AB=13\text{ cm},AC=5\text{ cm}$ — The question. AB is the diameter, and we want the third side BC.

2. 2

   $A{B}^2=A{C}^2+B{C}^2$ — Angle ACB is 90°, so use Pythagoras. The diameter is the hypotenuse, so it goes on its own.

3. 3

   $B{C}^2={13}^2-5^2$ — We want a shorter side, so subtract the smaller square from the bigger one.

4. 4

   $B{C}^2=169-25=144$ — Square each number first: 13² = 169 and 5² = 25.

5. 5

   $BC=\sqrt{144}=12\text{ cm}$ — Square root to undo the square. A length is always positive.

6. 6

   $\text{Area}=\frac{1}{2}\times 5\times 12=30{\text{ cm}}^2$ — The two shorter sides meet at the right angle, so they are the base and the height.

Worked example

1. 1

   $AC=7\text{ cm},BC=24\text{ cm}$ — The question. AB is a diameter. Find the radius of the circle.

2. 2

   $A{B}^2=7^2+24^2$ — The angle at C is 90°, so the diameter AB is the hypotenuse. To find the hypotenuse, add the squares.

3. 3

   $A{B}^2=49+576=625$ — Square each number first: 7² = 49 and 24² = 576.

4. 4

   $AB=\sqrt{625}=25\text{ cm}$ — Square root to undo the square. This is the diameter.

5. 5

   $r=25÷2=12.5\text{ cm}$ — The radius is half the diameter, because O is the midpoint of AB.

Worked example

1. 1

   $AC=12\text{ cm},BC=9\text{ cm}$ — The question. AB is a diameter. Find angle CAB.

2. 2

   $\angle ACB=90^{\circ }$ — Angle in a semicircle is 90°, so triangle ABC is right-angled at C.

3. 3

   $\tan (\angle CAB)=\frac{9}{12}$ — From A, BC is opposite and AC is adjacent, so use tan.

4. 4

   $\angle CAB={\tan }^{-1}(0.75)$ — 9 ÷ 12 = 0.75. Use inverse tan on your calculator to turn the ratio into an angle.

5. 5

   $\angle CAB=36.9^{\circ }$ — 36.869…°, rounded to 1 decimal place.

Worked example

1. 1

   $\angle P=34^{\circ },\angle Q=56^{\circ }$ — The question. P, Q and R are on a circle. Which side of triangle PQR is a diameter?

2. 2

   $\angle R=180^{\circ }-34^{\circ }-56^{\circ }$ — Angles in a triangle add up to 180°.

3. 3

   $\angle R=90^{\circ }$ — 180 − 34 = 146, then 146 − 56 = 90.

4. 4

   $PQ\text{is a diameter}$ — The 90° is at R, so the side opposite R, PQ, is the diameter.

5. 5

   $\text{centre}=\text{midpoint of }PQ$ — Every diameter passes through the centre, and the centre is halfway along it.