Sine Rule, Cosine Rule and Area of a Triangle

Angle PQR is at Q: which two sides meet at Q, and which side is opposite it?

1. Angle PQR is between PQ=5 and QR=8; the side opposite it is PR=10
2. Cosine rule: 10^2=5^2+8^2-2\times5\times8\times\cos PQR
3. Rearrange: \cos PQR=\dfrac{5^2+8^2-10^2}{2\times5\times8}
4. The first and fourth options are for angles PRQ and QPR; the second has the signs the wrong way round
5. Answer: \cos PQR=\dfrac{5^2+8^2-10^2}{2\times5\times8}

You know two sides and the angle between them, so use area =\frac12ab\sin C.

1. The 72^\circ angle is between the sides 6.4 cm and 9.5 cm
2. Area =\frac12\times6.4\times9.5\times\sin72^\circ
3. =30.4\times0.9510\ldots=28.912\ldots
4. Answer: 28.9 cm^2

Pair each side with the angle opposite it, then use the sine rule.

1. BC is opposite angle A (52^\circ) and AC is opposite angle B (71^\circ)
2. Sine rule: \dfrac{AC}{\sin 71^\circ}=\dfrac{9.6}{\sin 52^\circ}
3. AC=\dfrac{9.6\sin 71^\circ}{\sin 52^\circ}=11.518\ldots
4. Answer: 11.5 cm

Pair each side with the angle opposite it: DE is opposite angle F and EF is opposite angle D.

1. EF is opposite angle D and DE is opposite angle F
2. Sine rule: \dfrac{\sin D}{8.7}=\dfrac{\sin63^\circ}{11.2}
3. \sin D=\dfrac{8.7\sin63^\circ}{11.2}=0.6921\ldots
4. D=\sin^{-1}(0.6921\ldots)=43.798\ldots^\circ
5. EF is shorter than DE, so angle D is smaller than 63^\circ and cannot be obtuse
6. Answer: 43.8^\circ

Use the cosine rule, and remember that \cos120^\circ is negative.

1. \cos120^\circ=-\dfrac12
2. BC^2=7^2+8^2-2\times7\times8\times\left(-\dfrac12\right)
3. BC^2=49+64+56=169
4. BC=\sqrt{169}
5. Answer: 13 cm

Use area =\frac12ab\sin C; \sin135^\circ has the same value as \sin45^\circ.

1. \sin135^\circ=\sin(180^\circ-135^\circ)=\sin45^\circ=\dfrac{\sqrt2}{2}
2. Area =\frac12\times6\times10\times\sin135^\circ
3. =30\times\dfrac{\sqrt2}{2}
4. Answer: 15\sqrt2 cm^2

You know two sides and the angle between them, so use the cosine rule.

1. QR is opposite the 38^\circ angle
2. Cosine rule: QR^2=7.4^2+10.2^2-2\times 7.4\times 10.2\times\cos 38^\circ
3. QR^2=54.76+104.04-118.95\ldots=39.84\ldots
4. QR=\sqrt{39.84\ldots}=6.312\ldots
5. Answer: 6.31 cm

The smallest angle is opposite the shortest side; then use the cosine rule for an angle.

1. The shortest side is FG, so the smallest angle is opposite it: angle FHG
2. \cos FHG=\dfrac{8.1^2+9.4^2-5.8^2}{2\times8.1\times9.4}
3. =\dfrac{65.61+88.36-33.64}{152.28}=\dfrac{120.33}{152.28}=0.7901\ldots
4. Angle FHG=\cos^{-1}(0.7901\ldots)=37.796\ldots^\circ
5. Answer: 37.8^\circ

Put \cos A=\frac23 straight into the cosine rule; you do not need to find angle A.

1. BC^2=AB^2+AC^2-2\times AB\times AC\times\cos A
2. BC^2=81+64-2\times9\times8\times\dfrac23
3. 2\times9\times8\times\dfrac23=144\times\dfrac23=96
4. BC^2=145-96=49
5. Answer: BC=7 cm

BC is opposite the 45^\circ angle and AB is opposite the 120^\circ angle; use \sin120^\circ=\sin60^\circ.

1. Sine rule: \dfrac{BC}{\sin45^\circ}=\dfrac{12}{\sin120^\circ}
2. \sin45^\circ=\dfrac{\sqrt2}{2} and \sin120^\circ=\sin60^\circ=\dfrac{\sqrt3}{2}
3. BC=\dfrac{12\times\frac{\sqrt2}{2}}{\frac{\sqrt3}{2}}=\dfrac{12\sqrt2}{\sqrt3}
4. Multiply top and bottom by \sqrt3: \dfrac{12\sqrt6}{3}=4\sqrt6
5. Answer: 4\sqrt6 cm

The sine rule gives the acute angle with the same sine; the obtuse angle is 180^\circ minus it.

1. KM=13.6 is opposite x and KL=7.5 is opposite the 24^\circ angle
2. Sine rule: \dfrac{\sin x}{13.6}=\dfrac{\sin24^\circ}{7.5}
3. \sin x=\dfrac{13.6\sin24^\circ}{7.5}=0.7375\ldots
4. \sin^{-1}(0.7375\ldots)=47.523\ldots^\circ, which is acute
5. \sin(180^\circ-\theta)=\sin\theta, so x=180^\circ-47.523\ldots^\circ
6. x=132.476\ldots^\circ
7. Answer: x=132.5^\circ

No angle is given: find one angle with the cosine rule first, then use area =\frac12ab\sin C.

1. Find angle PQR (between PQ and QR) with the cosine rule
2. \cos PQR=\dfrac{7^2+9^2-12^2}{2\times7\times9}=\dfrac{-14}{126}=-\dfrac19
3. Angle PQR=\cos^{-1}\left(-\dfrac19\right)=96.379\ldots^\circ
4. Area =\frac12\times7\times9\times\sin96.379\ldots^\circ
5. =31.304\ldots
6. Answer: 31.3 cm^2

Call AC x, so AB=2x, and put both into area =\frac12ab\sin C.

1. Let AC=x cm, so AB=2x cm; angle BAC is between them
2. Area =\frac12\times2x\times x\times\sin38^\circ=x^2\sin38^\circ
3. x^2\sin38^\circ=52, so x^2=\dfrac{52}{\sin38^\circ}=84.461\ldots
4. x=\sqrt{84.461\ldots}=9.1903\ldots
5. Answer: AC=9.19 cm

A diagonal splits the rhombus into two identical triangles with two sides of 7 cm, so the area is 7\times7\times\sin\theta.

1. A diagonal splits the rhombus into two triangles, each with sides 7, 7 and angle \theta between them
2. Area =2\times\frac12\times7\times7\times\sin\theta=49\sin\theta
3. 49\sin\theta=40, so \sin\theta=\dfrac{40}{49}
4. \sin^{-1}\left(\dfrac{40}{49}\right)=54.718\ldots^\circ, the acute angle
5. Neighbouring angles of a rhombus add to 180^\circ: 180^\circ-54.718\ldots^\circ=125.281\ldots^\circ
6. Answer: 125.3^\circ

Sketch triangle CPQ with north lines at C and P, and use the bearings to find two of its angles.

1. At C: CQ points north and CP is on 040^\circ, so angle QCP=40^\circ
2. The bearing of C from P is 040^\circ+180^\circ=220^\circ
3. At P: angle CPQ=300^\circ-220^\circ=80^\circ
4. Angle CQP=180^\circ-40^\circ-80^\circ=60^\circ
5. Sine rule: \dfrac{CQ}{\sin80^\circ}=\dfrac{6.5}{\sin60^\circ}
6. CQ=\dfrac{6.5\sin80^\circ}{\sin60^\circ}=7.391\ldots
7. Answer: 7.39 km

Work out the length of each side from the coordinates first, then use the cosine rule. Answer part (a) before attempting part (b).

Part (a)

1. Use Pythagoras for each side
2. AB^2=6^2+2^2=40 BC^2=3^2+4^2=25 AC^2=3^2+6^2=45
3. Cosine rule: \cos ABC=\dfrac{AB^2+BC^2-AC^2}{2\times AB\times BC}
4. =\dfrac{40+25-45}{2\times\sqrt{40}\times5}=\dfrac{20}{10\sqrt{40}}=0.3162\ldots
5. Angle ABC=\cos^{-1}(0.3162\ldots)=71.565\ldots^\circ
6. Answer: 71.6^\circ

Part (b)

1. Area =\frac12\times AB\times BC\times\sin ABC
2. =\frac12\times\sqrt{40}\times5\times\sin71.565\ldots^\circ
3. =15.0
4. Answer: 15 units^2

Use the cosine rule for BC with \cos60^\circ=\frac12, then expand and collect into a quadratic. Answer part (a) before attempting part (b).

Part (a)

1. Cosine rule: 13^2=x^2+(x+7)^2-2x(x+7)\cos60^\circ
2. \cos60^\circ=\dfrac12, so the last term is -x(x+7)
3. 169=x^2+x^2+14x+49-x^2-7x
4. 169=x^2+7x+49, so x^2+7x-120=0
5. (x+15)(x-8)=0, so x=-15 or x=8
6. x is a length, so it is positive
7. Answer: x=8

Part (b)

1. The sides are AB=8, AC=15 and BC=13
2. The largest angle is opposite the longest side, AC: angle ABC
3. \cos ABC=\dfrac{8^2+13^2-15^2}{2\times8\times13}=\dfrac{8}{208}
4. Angle ABC=\cos^{-1}\left(\dfrac{8}{208}\right)=87.795\ldots^\circ
5. Answer: 87.8^\circ

Use \sin^2A+\cos^2A=1 to find \sin BAC; angle BAC is obtuse, but its sine is still positive. Answer part (a) before attempting part (b).

Part (a)

1. \sin^2 A+\cos^2 A=1
2. \sin^2 A=1-\dfrac{9}{25}=\dfrac{16}{25}
3. Angle A is between 0^\circ and 180^\circ, so \sin A is positive
4. Answer: \sin BAC=\dfrac45

Part (b)

1. Area =\frac12\times6\times10\times\sin BAC
2. =30\times\dfrac45
3. Answer: 24 cm^2

Part (c)

1. BC^2=6^2+10^2-2\times6\times10\times\left(-\dfrac35\right)
2. =36+100+72=208
3. BC=\sqrt{208}=\sqrt{16\times13}
4. Answer: 4\sqrt{13} cm

Find AC first and keep its full calculator value; then split the quadrilateral into two triangles along AC.

Part (a)

1. In triangle ABC you know two sides and the angle between them: use the cosine rule
2. AC^2=8^2+11^2-2\times 8\times 11\times\cos 110^\circ
3. \cos 110^\circ is negative, so AC^2=64+121+60.19\ldots=245.19\ldots
4. AC=15.658\ldots
5. Answer: 15.7 cm

Part (b)

1. Area of triangle ABC=\frac12\times 8\times 11\times\sin 110^\circ=41.346\ldots
2. In triangle ACD: angle ADC=180^\circ-40^\circ-65^\circ=75^\circ
3. Sine rule: \dfrac{AD}{\sin 65^\circ}=\dfrac{AC}{\sin 75^\circ}, so AD=\dfrac{15.658\ldots\times\sin 65^\circ}{\sin 75^\circ}=14.692\ldots
4. Area of triangle ACD=\frac12\times 15.658\ldots\times 14.692\ldots\times\sin 40^\circ=73.940\ldots
5. Total area =41.346\ldots+73.940\ldots=115.28\ldots
6. Answer: 115 cm^2

Use the cosine rule with exact values for PR; then the angle POR at the centre is twice angle PQR. Answer part (a) before attempting part (b).

Part (a)

1. Cosine rule: PR^2=(3\sqrt2)^2+7^2-2\times3\sqrt2\times7\times\cos45^\circ
2. (3\sqrt2)^2=18 and \cos45^\circ=\dfrac{\sqrt2}{2}
3. 2\times3\sqrt2\times7\times\dfrac{\sqrt2}{2}=21\times2=42
4. PR^2=18+49-42=25
5. Answer: PR=5 cm

Part (b)

1. The angle at the centre is twice the angle at the circumference
2. Angle POR=2\times45^\circ=90^\circ
3. OP=OR=r, so in right-angled triangle POR: r^2+r^2=PR^2=25
4. r^2=\dfrac{25}{2}
5. Area =\pi r^2
6. Answer: \dfrac{25}{2}\pi cm^2