Exact Circular-Measure Problems | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Use exact triangle values

An equilateral triangle has three equal sides and three equal angles. Its angles add to 180°, so each is 60°. Split one of side 2 down the middle: each right triangle has hypotenuse 2, short side 1 and height $\sqrt{2^2-1^2}=\sqrt3$ . The smaller angle is 30°, or $\pi/6$ .

Sine is opposite/hypotenuse; cosine is adjacent/hypotenuse. The adjacent side is the side next to the angle, excluding the hypotenuse. Therefore $\sin(\pi/6)=1/2$ and $\cos(\pi/6)=\sqrt3/2$ .

When the triangle has unequal sides

If you know two sides $a$ and $b$ and the angle $C$ between them, the finds the opposite side $c$ :

$c^2=a^2+b^2-2ab\cos C$ .

This extends Pythagoras to a triangle without a right angle. Split side $b$ into a horizontal part $b\cos C$ and height $b\sin C$ . Pythagoras then gives:

$c^2=(a-b\cos C)^2+(b\sin C)^2$ .

Expand the squares and use $\sin^2C+\cos^2C=1$ to obtain the rule. Here $\sin^2C$ means $(\sin C)^2$ . This equation is true for every angle; it follows from Pythagoras with hypotenuse 1. For an obtuse angle, the horizontal part points left, so its cosine is negative; squaring removes the sign and the rule still works.

Worked example

Show that BC = r√(5 − 2√3)

A sector has radius $OA=OB=2r$ , angle $\angle AOB=\pi/6$ , and C is the midpoint of OA. Hence $OC=r$ . (9709/12 May/Jun 2020 Q7(a))

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Triangle OBC has sides r and 2r enclosing π/6.

In triangle OBC, the known sides are $2r$ and $r$ , not two equal radii. Use $BC^2=(2r)^2+r^2-2(2r)(r)\cos(\pi/6)$ .
Insert the exact cosine: $BC^2=5r^2-4r^2(\sqrt3/2)$ .
Factor out $r^2$ : $BC^2=r^2(5-2\sqrt3)$ . Since $r>0$ and BC is a length, take the positive square root.

Answer

BC=r\sqrt{5-2\sqrt3}

The cosine rule also works for a chord, giving $AB^2=2r^2(1-\cos\theta)$ . For equal radii, the half-angle method is usually shorter; for unequal sides, use the cosine rule.

Keep the area exact too

Your turn9709/11 May/Jun 2022 Q5(a) [3 marks]

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The shaded region is sector ABC minus right triangle ABD.

Sector ABC has centre A, radius $r$ and angle $\angle CAB=\pi/6$ . D lies on AC and $BD\perp AC$ . Find the exact area between arc BC and the lines BD and DC, in terms of $r$ .

Show worked answer

Subtract right triangle ABD from the sector. The sector area is $\pi r^2/12$ . The right triangle has $BD=r\sin(\pi/6)=r/2$ and $AD=r\cos(\pi/6)=\sqrt3r/2$ , so its area is $\sqrt3r^2/8$ .

Answer

A=\frac{\pi r^2}{12}-\frac{\sqrt3r^2}{8}

If you want a single fraction, use denominator 24: $A=r^2(2\pi-3\sqrt3)/24$ .

Use exact triangle values

Sine is opposite/hypotenuse; cosine is adjacent/hypotenuse. The adjacent side is the side next to the angle, excluding the hypotenuse. Therefore $\sin(\pi/6)=1/2$ and $\cos(\pi/6)=\sqrt3/2$ .

When the triangle has unequal sides

If you know two sides $a$ and $b$ and the angle $C$ between them, the finds the opposite side $c$ :

$c^2=a^2+b^2-2ab\cos C$ .

This extends Pythagoras to a triangle without a right angle. Split side $b$ into a horizontal part $b\cos C$ and height $b\sin C$ . Pythagoras then gives:

$c^2=(a-b\cos C)^2+(b\sin C)^2$ .

Worked example

Show that BC = r√(5 − 2√3)

A sector has radius $OA=OB=2r$ , angle $\angle AOB=\pi/6$ , and C is the midpoint of OA. Hence $OC=r$ . (9709/12 May/Jun 2020 Q7(a))

Swipe left or right to see the whole diagram →

Triangle OBC has sides r and 2r enclosing π/6.

In triangle OBC, the known sides are $2r$ and $r$ , not two equal radii. Use $BC^2=(2r)^2+r^2-2(2r)(r)\cos(\pi/6)$ .
Insert the exact cosine: $BC^2=5r^2-4r^2(\sqrt3/2)$ .
Factor out $r^2$ : $BC^2=r^2(5-2\sqrt3)$ . Since $r>0$ and BC is a length, take the positive square root.

Answer

BC=r\sqrt{5-2\sqrt3}

The cosine rule also works for a chord, giving $AB^2=2r^2(1-\cos\theta)$ . For equal radii, the half-angle method is usually shorter; for unequal sides, use the cosine rule.

Keep the area exact too

Your turn9709/11 May/Jun 2022 Q5(a) [3 marks]

Swipe left or right to see the whole diagram →

The shaded region is sector ABC minus right triangle ABD.

Sector ABC has centre A, radius $r$ and angle $\angle CAB=\pi/6$ . D lies on AC and $BD\perp AC$ . Find the exact area between arc BC and the lines BD and DC, in terms of $r$ .

Show worked answer

Subtract right triangle ABD from the sector. The sector area is $\pi r^2/12$ . The right triangle has $BD=r\sin(\pi/6)=r/2$ and $AD=r\cos(\pi/6)=\sqrt3r/2$ , so its area is $\sqrt3r^2/8$ .

Answer

A=\frac{\pi r^2}{12}-\frac{\sqrt3r^2}{8}

If you want a single fraction, use denominator 24: $A=r^2(2\pi-3\sqrt3)/24$ .