C4.7; E4.7 - Right angles in circles

C4.7; E4.7 - Right angles in circles

A circle can guarantee a right angle without a protractor. You will use a diameter to identify an angle in a semicircle, use the right angle between a tangent and its radius, and combine these facts with triangle angles while giving precise reasons.

A right angle in a semicircle

Fix AA and BB at the ends of a diameter, then move CC around one semicircular arc. The triangle changes shape. Does its angle at CC change too? Surprisingly, that angle stays at 90∘90^\circ, provided CC is a different point on the circumference.

Angle in a semicircle

If ABAB is a diameter and CC is another point on the circle, then

∠ACB=90∘.\angle ACB=90^\circ.

The right angle is at the circumference, opposite the diameter. The reason is “angle in a semicircle”.

Not given in the List of formulas: learn this circle theorem.

Why? Join the centre OO to CC. The two smaller triangles have equal radii as sides, so each has two equal base angles. Call these base angles aa and bb. The large triangle's three angles are aa, bb and a+ba+b. Their total gives 2(a+b)=180∘2(a+b)=180^\circ, so a+b=90∘a+b=90^\circ. The moving point changes aa and bb, but not their sum.

[DIAGRAM: asset_name: A diameter fixes the opposite right angle; asset_slug: semicircle_right_angles; caption: C moves, but angle ACB stays a right angle because AB is a diameter.; description: Three exact circles have horizontal diameter AB through centre O and a point C on the upper circumference. The interior angle CAB is respectively 32, 57 and 41 degrees. Angle ACB is marked as a right angle and ABC is labelled x, y or z. In the first circle dashed radius OC illustrates the two isosceles triangles.]
C moves, but angle ACB stays a right angle because AB is a diameter.

Worked example 1: an angle beside a diameter

Question. ABAB is a diameter, CC lies on the circle, and ∠CAB=32∘\angle CAB=32^\circ. Calculate ∠ABC\angle ABC, giving geometric reasons.

Plan. Establish the right angle at CC, then complete the triangle.

∠ACB=90∘angle in a semicircle∠ABC=180∘−90∘−32∘triangle angle sum=58∘\begin{aligned} \angle ACB&=90^\circ && \text{angle in a semicircle}\\ \angle ABC&=180^\circ-90^\circ-32^\circ && \text{triangle angle sum}\\ &=58^\circ \end{aligned}

Answer. ∠ABC=58∘\angle ABC=58^\circ.

Check. The two acute angles in this right-angled triangle add to 32+58=90∘32+58=90^\circ.

Move CC so that the given angle changes. The diameter and the right angle remain.

Worked example 2: a different point on the arc

Question. ABAB is still a diameter and CC still lies on the circle, but ∠CAB=57∘\angle CAB=57^\circ. Find ∠ABC\angle ABC, with reasons.

Plan. Keep the semicircle right angle and subtract the new known angle.

∠ACB=90∘angle in a semicircle∠ABC=180∘−90∘−57∘triangle angle sum=33∘\begin{aligned} \angle ACB&=90^\circ && \text{angle in a semicircle}\\ \angle ABC&=180^\circ-90^\circ-57^\circ && \text{triangle angle sum}\\ &=33^\circ \end{aligned}

Answer. ∠ABC=33∘\angle ABC=33^\circ.

Check. 57+33=9057+33=90. Increasing the angle at AA by 25∘25^\circ has decreased the angle at BB by 25∘25^\circ.

The word diameter makes the angle exact. A drawing that merely looks right-angled is weaker evidence.

Watch out

“A triangle inside a circle is right-angled” is false. You need one side to be a diameter, not just a chord. The right angle is opposite that diameter, not at its midpoint.

Use the diameter condition in your written reason.

A tangent meets its radius at a right angle

A tangent touches the circle at one point. Join that point of contact to the centre. That particular radius is perpendicular to the tangent: the angle is 90∘90^\circ, whichever way the diagram is turned.

Tangent and radius

If a line is tangent to a circle with centre OO at TT, then

OT⊥tangent at T.OT\perp\text{tangent at }T.

The radius must meet the tangent at its point of contact. A radius to another point on the circle does not give this conclusion.

Not given in the List of formulas: learn this circle theorem.

A line touching a circle is different from a line cutting through it. The right-angle fact supplies exact information from the word tangent; it does not come from estimating a nearly square-looking corner.

[DIAGRAM: asset_name: Radius and tangent form a right angle; asset_slug: tangent_radius_triangles; caption: The right angle is at T, where radius OT meets the tangent TP.; description: Three circles with centre O have tangent TP at T and an external point P on that tangent. OP forms a triangle with radius OT and tangent segment TP. Angle TOP is respectively 35, 62 and 47 degrees. The angle at T is marked right and OPT is labelled x, y or z. The circle, tangent and triangle coordinates agree exactly with each given angle.]
The right angle is at T, where radius OT meets the tangent TP.

Worked example 3: a triangle outside the circle

Question. PTPT is tangent at TT to a circle with centre OO, and ∠TOP=35∘\angle TOP=35^\circ. Find ∠OPT\angle OPT, giving reasons.

Plan. Use the tangent–radius right angle at TT, then use triangle OTPOTP.

∠OTP=90∘radius perpendicular to tangent∠OPT=180∘−90∘−35∘triangle angle sum=55∘\begin{aligned} \angle OTP&=90^\circ && \text{radius perpendicular to tangent}\\ \angle OPT&=180^\circ-90^\circ-35^\circ && \text{triangle angle sum}\\ &=55^\circ \end{aligned}

Answer. ∠OPT=55∘\angle OPT=55^\circ.

Check. 35+55=9035+55=90, so the two non-right angles complement one another.

Continue this lesson.

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