The Cosine Rule

The known angle is between the two known sides.

1. The known angle 62^\circ is between the two known sides, so use the cosine rule: a^2=b^2+c^2-2bc\cos A.
2. BC^2=5^2+8^2-2\times5\times8\cos62^\circ.
3. This gives BC^2, so take the square root to get BC.
4. The first option forgets the square root; the last has + instead of -; and \sqrt{5^2+8^2} is Pythagoras, which needs a right angle.

Use the two sides enclosing the known angle.

1. \angle BAC=48^\circ is between the known sides AB=9 and AC=6, so use the cosine rule: BC^2=AB^2+AC^2-2\times AB\times AC\cos A.
2. BC^2=9^2+6^2-2\times9\times6\cos48^\circ.
3. BC^2=117-72.266\ldots=44.733\ldots
4. BC=\sqrt{44.733\ldots}=6.688\ldots, which is 6.7 cm to 1 decimal place.

The cosine rule also works when the included angle is obtuse.

1. \angle BAC=123^\circ is between AB=11 and AC=7, so use the cosine rule: BC^2=11^2+7^2-2\times11\times7\cos123^\circ.
2. \cos123^\circ is negative, so -2\times11\times7\cos123^\circ is +83.874\ldots
3. BC^2=170+83.874\ldots=253.874\ldots
4. BC=\sqrt{253.874\ldots}=15.933\ldots, which is 15.9 cm to 1 decimal place.

The side opposite the required angle is the one subtracted in the numerator.

1. All three sides are known, so use the cosine rule for an angle: \cos A=\dfrac{b^2+c^2-a^2}{2bc}, where a is the side opposite A.
2. The side opposite \angle BAC is BC=8, so \cos A=\dfrac{12^2+15^2-8^2}{2\times12\times15}=\dfrac{305}{360}=0.84722\ldots
3. \angle BAC=\cos^{-1}(0.84722\ldots)=32.089\ldots^\circ, which is 32.1^\circ to 1 decimal place.

Isolate the cosine term before dividing.

1. Add 2ac\cos B to both sides and subtract b^2: 2ac\cos B=a^2+c^2-b^2.
2. Divide by 2ac: \cos B=\dfrac{a^2+c^2-b^2}{2ac}.
3. The side opposite the angle, b, is the one subtracted. The first option has every sign the wrong way round.

The largest angle is opposite the longest side.

1. The largest angle is opposite the longest side, 13 cm. Call it \theta.
2. Cosine rule: \cos\theta=\dfrac{8^2+10^2-13^2}{2\times8\times10}=\dfrac{-5}{160}=-0.03125.
3. The cosine is negative, so the angle is obtuse: \theta=\cos^{-1}(-0.03125)=91.790\ldots^\circ.
4. The largest angle is 91.8^\circ to 1 decimal place.

Use the angle between the equal sides.

1. The angle 48^\circ is between the two 7 cm sides, so use the cosine rule for the third side x.
2. x^2=7^2+7^2-2\times7\times7\cos48^\circ=98-65.574\ldots=32.425\ldots
3. x=\sqrt{32.425\ldots}=5.694\ldots, which is 5.7 cm to 1 decimal place.

Convert the lengths to the same unit before using the cosine rule.

1. Use the same unit for every length: AB=90 mm =9 cm.
2. \angle BAC=63^\circ is between AB and AC, so use the cosine rule: BC^2=9^2+12^2-2\times9\times12\cos63^\circ.
3. BC^2=225-98.061\ldots=126.938\ldots
4. BC=\sqrt{126.938\ldots}=11.266\ldots, which is 11.3 cm to 1 decimal place.

The angle is between the two known sides. Use the exact value of \cos60^\circ.

1. \angle BAC=60^\circ is between AB and AC, so use the cosine rule: BC^2=8^2+5^2-2\times8\times5\cos60^\circ.
2. \cos60^\circ=\dfrac12, so 2\times8\times5\times\dfrac12=40.
3. BC^2=64+25-40=49.
4. BC=\sqrt{49}=7 cm.

Use the two sides meeting at the required angle in the denominator.

1. The side opposite \angle PQR is PR=10.6; the sides meeting at Q are PQ=12.1 and QR=7.8.
2. Cosine rule: \cos Q=\dfrac{12.1^2+7.8^2-10.6^2}{2\times12.1\times7.8}=\dfrac{94.89}{188.76}=0.50270\ldots
3. \angle PQR=\cos^{-1}(0.50270\ldots)=59.821\ldots^\circ, which is 59.8^\circ to 1 decimal place.

The two distances from the gate enclose the given angle.

1. The angle at the gate, 76^\circ, is between the two known distances, so use the cosine rule for BF.
2. BF^2=51^2+34^2-2\times51\times34\cos76^\circ.
3. BF^2=3757-838.98\ldots=2918.01\ldots
4. BF=\sqrt{2918.01\ldots}=54.018\ldots, which is 54.0 m to 1 decimal place.

Find the missing side before adding the three lengths.

1. The angle 73^\circ is between the sides 10.5 cm and 8.4 cm, so use the cosine rule for the third side x.
2. x^2=10.5^2+8.4^2-2\times10.5\times8.4\cos73^\circ=129.235\ldots
3. x=\sqrt{129.235\ldots}=11.368\ldots cm.
4. Perimeter =10.5+8.4+11.368\ldots=30.268\ldots, which is 30.3 cm to 1 decimal place.

Find an angle in triangle ABC, then use the right-angled triangle containing AH.

1. Triangle ABH is right-angled at H, so \angle BAH=90^\circ-\angle ABC. First find \angle ABC.
2. All three sides of ABC are known. The side opposite B is AC=9.2, so \cos B=\dfrac{14.1^2+17.3^2-9.2^2}{2\times14.1\times17.3}=0.84749\ldots
3. \angle ABC=\cos^{-1}(0.84749\ldots)=32.059\ldots^\circ.
4. \angle BAH=90^\circ-32.059\ldots^\circ=57.940\ldots^\circ, which is 57.9^\circ to 1 decimal place.

Find the interior angle at B, then use angles on a straight line.

1. First find the interior angle \angle ABC. The side opposite B is AC=11.
2. Cosine rule: \cos B=\dfrac{14^2+8^2-11^2}{2\times14\times8}=\dfrac{139}{224}=0.62053\ldots
3. \angle ABC=\cos^{-1}(0.62053\ldots)=51.644\ldots^\circ.
4. Angles on a straight line add to 180^\circ: x=180^\circ-51.644\ldots^\circ=128.355\ldots^\circ, which is 128.4^\circ to 1 decimal place.

Find a diagonal, then use the fact that the diagonals of a parallelogram bisect each other.

1. BO is half of the diagonal BD, so find BD first.
2. In triangle ABD, the angle 58^\circ is between AB=14 and AD=9. Cosine rule: BD^2=14^2+9^2-2\times14\times9\cos58^\circ=143.460\ldots
3. BD=\sqrt{143.460\ldots}=11.977\ldots cm.
4. The diagonals of a parallelogram bisect each other, so BO=11.977\ldots\div2=5.988\ldots, which is 6.0 cm to 1 decimal place.

Square the surd length, then compare the cosine with an exact trigonometric value.

1. All three sides are known. The side opposite \angle BAC is BC=2\sqrt7, and (2\sqrt7)^2=4\times7=28.
2. Cosine rule: \cos A=\dfrac{6^2+4^2-28}{2\times6\times4}=\dfrac{24}{48}=\dfrac12.
3. \cos60^\circ=\dfrac12, so \angle BAC=60^\circ.

Find an angle at the base, then use the sine ratio in a right-angled triangle.

1. The height from C makes a right-angled triangle with hypotenuse AC=10 and angle A, so find \angle BAC first.
2. The side opposite A is BC=14: \cos A=\dfrac{12^2+10^2-14^2}{2\times12\times10}=\dfrac{48}{240}=0.2.
3. \angle BAC=\cos^{-1}(0.2)=78.463\ldots^\circ.
4. Height =AC\sin A=10\sin78.463\ldots^\circ=9.797\ldots, which is 9.8 cm to 1 decimal place.

Find the part of the angle at C in triangle ABC before using triangle ACD.

1. In triangle ACD we know AC=8 and CD=5, but not the angle \angle ACD between them. It is part of \angle BCD=145^\circ.
2. Triangle ABC has three known sides. The side opposite \angle ACB is AB=11: \cos(\angle ACB)=\dfrac{8^2+7^2-11^2}{2\times8\times7}=\dfrac{-8}{112}, so \angle ACB=94.096\ldots^\circ.
3. \angle ACD=145^\circ-94.096\ldots^\circ=50.903\ldots^\circ.
4. Cosine rule in triangle ACD: AD^2=8^2+5^2-2\times8\times5\cos50.903\ldots^\circ=38.550\ldots
5. AD=\sqrt{38.550\ldots}=6.208\ldots, which is 6.2 cm to 1 decimal place.

First use the equal sides of a rhombus to find the sides of triangle ABC.

1. A rhombus has four equal sides, so each side is 44\div4=11 cm.
2. Triangle ABC has AB=BC=11 and AC=17. The side opposite \angle ABC is AC.
3. Cosine rule: \cos B=\dfrac{11^2+11^2-17^2}{2\times11\times11}=\dfrac{-47}{242}=-0.19421\ldots
4. \angle ABC=\cos^{-1}(-0.19421\ldots)=101.198\ldots^\circ, which is 101.2^\circ to 1 decimal place.

The required angle is the sum of two angles at A.

1. \angle BAD=\angle BAC+\angle CAD. Each triangle has three known sides, so use the cosine rule in each.
2. Triangle ABC: the side opposite \angle BAC is BC=8, so \cos(\angle BAC)=\dfrac{13^2+11^2-8^2}{2\times13\times11}=\dfrac{226}{286} and \angle BAC=37.794\ldots^\circ.
3. Triangle ACD: the side opposite \angle CAD is CD=7, so \cos(\angle CAD)=\dfrac{10^2+11^2-7^2}{2\times10\times11}=\dfrac{172}{220} and \angle CAD=38.572\ldots^\circ.
4. \angle BAD=37.794\ldots^\circ+38.572\ldots^\circ=76.367\ldots^\circ, which is 76.4^\circ to 1 decimal place.