Mixed Sine and Cosine Rule Problems

The given angle lies between the two known sides.

1. No side is known together with its opposite angle, so the sine rule cannot be used.
2. The angle 52^\circ is between the two known sides, so use the cosine rule: x^2=6^2+9^2-2\times6\times9\cos52^\circ.
3. The cosine rule uses \cos, not \sin; and x^2=6^2+9^2 is Pythagoras, which needs a right angle.

Consider which rule can find an angle when all three sides are known.

1. Area =\dfrac12ab\sin C needs two sides and the angle between them, but no angle is known.
2. With all three sides known, the cosine rule finds an angle.
3. Then use \dfrac12ab\sin C with the two sides that enclose that angle. Pythagoras does not apply, as 7^2+10^2\neq12^2, so there is no right angle.

Find the third side before working out how much wire is used.

1. The angle 57^\circ is between the sides 6.5 and 9.2, so the cosine rule gives the third side x.
2. x^2=6.5^2+9.2^2-2\times6.5\times9.2\cos57^\circ=61.751\ldots, so x=7.858\ldots cm.
3. Wire used =6.5+9.2+7.858\ldots=23.558\ldots cm.
4. Wire left over =30-23.558\ldots=6.441\ldots, which is 6.4 cm to 1 decimal place.

Find the full side length before using the midpoint.

1. The known side AB=11 is opposite \angle C=180^\circ-43^\circ-68^\circ=69^\circ.
2. AC is opposite \angle B=68^\circ. Sine rule: AC=\dfrac{11\sin68^\circ}{\sin69^\circ}=10.924\ldots cm.
3. M is the midpoint of AC, so AM=10.924\ldots\div2=5.462\ldots
4. AM=5.5 cm to 1 decimal place.

Consider the known information in each triangle in turn.

1. In triangle ABC, the two angles give the third angle, \angle ACB, which is opposite AB. The sine rule then finds AC.
2. In triangle ACD, the known angle 47^\circ lies between AC and AD, so the cosine rule finds CD.
3. The cosine rule cannot come first, because triangle ABC has only one known side.

Recover the included angle from the area before finding the third side.

1. AB and AC enclose \angle BAC, so 16=\dfrac12\times6\times9\times\sin A=27\sin A.
2. \sin A=\dfrac{16}{27}, and the acute angle is A=\sin^{-1}\left(\dfrac{16}{27}\right)=36.341\ldots^\circ.
3. Now two sides and the angle between them are known, so use the cosine rule: BC^2=6^2+9^2-2\times6\times9\cos36.341\ldots^\circ=30.005\ldots
4. BC=\sqrt{30.005\ldots}=5.477\ldots, which is 5.5 cm to 1 decimal place.

The inverse-sine result is acute, so use its supplement for the stated obtuse angle.

1. AB and AC enclose \angle BAC: 30=\dfrac12\times7\times12\times\sin A=42\sin A, so \sin A=\dfrac{30}{42}=\dfrac57.
2. \sin^{-1}\left(\dfrac57\right)=45.584\ldots^\circ is acute. The obtuse angle with the same sine is 180^\circ-45.584\ldots^\circ=134.415\ldots^\circ.
3. Cosine rule: BC^2=7^2+12^2-2\times7\times12\cos134.415\ldots^\circ=310.575\ldots (the cosine is negative, so the last term adds on).
4. BC=\sqrt{310.575\ldots}=17.623\ldots, which is 17.6 cm to 1 decimal place.

Find the third side before choosing a rule to find the required angle.

1. Two sides and the angle between them are known, so first use the cosine rule for BC, the side opposite 58^\circ: BC^2=7.5^2+10.2^2-2\times7.5\times10.2\cos58^\circ=79.212\ldots
2. BC=8.900\ldots cm. Keep this value in your calculator.
3. Now BC and its opposite angle are known, so use the sine rule. AB=7.5 is opposite \angle ACB: \sin C=\dfrac{7.5\sin58^\circ}{8.900\ldots}=0.71463\ldots
4. \angle ACB=\sin^{-1}(0.71463\ldots)=45.613\ldots^\circ. The obtuse option, 134.4^\circ, is impossible as 58^\circ+134.4^\circ>180^\circ. So \angle ACB=45.6^\circ to 1 decimal place.

Use the non-right-angled triangle to find the shared side first.

1. In triangle ABC the angle 57^\circ is between AB=11 and AC=8, so the cosine rule gives the shared side BC.
2. BC^2=11^2+8^2-2\times11\times8\cos57^\circ=89.143\ldots
3. Triangle BCD has a right angle at B, so by Pythagoras CD^2=BC^2+BD^2=89.143\ldots+6^2=125.143\ldots
4. CD=\sqrt{125.143\ldots}=11.186\ldots, which is 11.2 cm to 1 decimal place.

Find both the base length and the area before using the base-height formula.

1. Area =\dfrac12\times BC\times h, where h is the height from A to BC. So find the area and BC.
2. Area =\dfrac12\times10\times14\times\sin63^\circ=62.370\ldots cm^2.
3. Cosine rule: BC^2=10^2+14^2-2\times10\times14\cos63^\circ=168.882\ldots, so BC=12.995\ldots cm.
4. The new perpendicular is the height corresponding to the derived base BC.

5. h=\dfrac{2\times\text{area}}{BC}=\dfrac{2\times62.370\ldots}{12.995\ldots}=9.598\ldots
6. h=9.6 cm to 1 decimal place.

Find the diagonal using the triangle with two known angles.

1. Triangle ABC has two known angles and a known side, so use the sine rule for the diagonal AC.
2. AC is opposite \angle ABC=80^\circ and AB=14 is opposite \angle BCA=65^\circ: AC=\dfrac{14\sin80^\circ}{\sin65^\circ}=15.212\ldots cm.
3. Triangle ACD now has three known sides. The side opposite \angle ADC is AC.
4. Cosine rule: \cos D=\dfrac{12^2+10^2-(15.212\ldots)^2}{2\times12\times10}=0.05240\ldots
5. \angle ADC=\cos^{-1}(0.05240\ldots)=86.996\ldots^\circ, which is 87.0^\circ to 1 decimal place.

The cosine rule can be used in each triangle on either side of the diagonal.

1. In triangle ABC, \angle ABC=64^\circ is between AB=11 and BC=7, so the cosine rule gives the diagonal: AC^2=11^2+7^2-2\times11\times7\cos64^\circ=102.490\ldots
2. AC=10.123\ldots cm.
3. In triangle ACD, \angle CAD=52^\circ is between AC and AD=8, so use the cosine rule again: CD^2=(10.123\ldots)^2+8^2-2\times10.123\ldots\times8\cos52^\circ=66.765\ldots
4. CD=\sqrt{66.765\ldots}=8.171\ldots, which is 8.2 cm to 1 decimal place.

The diagonal is a side of both triangles.

1. Triangle ABC has two known angles and a known side, so use the sine rule for the diagonal AC.
2. AC is opposite \angle ABC=74^\circ and AB=12 is opposite \angle BCA=52^\circ: AC=\dfrac{12\sin74^\circ}{\sin52^\circ}=14.638\ldots cm.
3. In triangle ACD, the angle \angle CAD=104^\circ is between AC and AD=9, so use the cosine rule.
4. CD^2=(14.638\ldots)^2+9^2-2\times14.638\ldots\times9\cos104^\circ=359.023\ldots
5. CD=\sqrt{359.023\ldots}=18.947\ldots, which is 18.9 cm to 1 decimal place.

Find the other side enclosing the known angle from the area formula.

1. AB and AC enclose 42^\circ, so use the area to find AC: 54=\dfrac12\times12\times AC\times\sin42^\circ.
2. AC=\dfrac{108}{12\sin42^\circ}=13.450\ldots cm.
3. Now use the cosine rule: BC^2=12^2+(13.450\ldots)^2-2\times12\times13.450\ldots\cos42^\circ=85.017\ldots
4. BC=\sqrt{85.017\ldots}=9.220\ldots, which is 9.2 cm to 1 decimal place.

Use the cosine rule to find an angle between two known sides, then use the triangle area formula.

1. All three sides are known, so use the cosine rule to find A.
2. \cos A=\dfrac{8^2+11^2-14^2}{2\times8\times11}=-\dfrac{1}{16}.
3. A=93.583321\ldots^\circ.
4. The sides enclosing A are 8 m and 11 m, so \text{area}=\dfrac12\times8\times11\times\sin(93.583321\ldots^\circ).
5. The area is 43.9 m^2 to 1 decimal place.

Recover the shared diagonal from the area, then use all three sides of the second triangle.

1. In triangle ABC, AB and AC enclose 48^\circ, so use the area: 42=\dfrac12\times12\times AC\times\sin48^\circ.
2. AC=\dfrac{84}{12\sin48^\circ}=9.419\ldots cm.
3. Triangle ACD now has three known sides. The side opposite \angle ADC is AC: \cos D=\dfrac{8^2+6^2-(9.419\ldots)^2}{2\times8\times6}=0.11744\ldots
4. \angle ADC=\cos^{-1}(0.11744\ldots)=83.255\ldots^\circ, which is 83.3^\circ to 1 decimal place.

Find the angle at C in the first triangle and use the straight line to find the angle in the second.

1. Triangle ABC has three known sides. The side opposite \angle ACB is AB=13: \cos(\angle ACB)=\dfrac{9^2+10^2-13^2}{2\times9\times10}=\dfrac{12}{180}, so \angle ACB=86.177\ldots^\circ.
2. BCD is a straight line, so \angle ACD=180^\circ-86.177\ldots^\circ=93.822\ldots^\circ.
3. Use the straight line at C to carry the angle into triangle ACD.

4. In triangle ACD, AD is opposite \angle ACD and AC=10 is opposite \angle ADC=35^\circ. Sine rule: \dfrac{AD}{\sin93.822\ldots^\circ}=\dfrac{10}{\sin35^\circ}.
5. AD=\dfrac{10\sin93.822\ldots^\circ}{\sin35^\circ}=17.395\ldots, which is 17.4 cm to 1 decimal place.

Find the angle at A in the first triangle so that you can split the given interior angle.

1. In triangle ACD we know AC=13 and AD=8; we need \angle CAD, which is part of \angle BAD=123^\circ. So find \angle BAC first.
2. Triangle ABC: AC=13 is opposite \angle ABC=76^\circ and AB=9 is opposite \angle ACB. Sine rule: \sin C=\dfrac{9\sin76^\circ}{13}=0.67174\ldots, so \angle ACB=42.201\ldots^\circ (acute, as AB<AC).
3. Angles in triangle ABC: \angle BAC=180^\circ-76^\circ-42.201\ldots^\circ=61.798\ldots^\circ. So \angle CAD=123^\circ-61.798\ldots^\circ=61.201\ldots^\circ.
4. Subtract the first triangle angle from the whole interior angle to find the second triangle angle.

5. Cosine rule in triangle ACD: CD^2=13^2+8^2-2\times13\times8\cos61.201\ldots^\circ=132.800\ldots
6. CD=\sqrt{132.800\ldots}=11.523\ldots, which is 11.5 cm to 1 decimal place.

Find each part of the angle at A, choosing the rule that matches the available information.

1. \angle BAD=\angle BAC+\angle CAD. Find each part.
2. Triangle ABC: AC=14 is opposite \angle ABC=56^\circ and AB=10 is opposite \angle ACB. Sine rule: \sin C=\dfrac{10\sin56^\circ}{14}=0.59216\ldots, so \angle ACB=36.311\ldots^\circ (acute, as AB<AC).
3. So \angle BAC=180^\circ-56^\circ-36.311\ldots^\circ=87.688\ldots^\circ.
4. Triangle ACD has three known sides. The side opposite \angle CAD is CD=9: \cos(\angle CAD)=\dfrac{14^2+12^2-9^2}{2\times14\times12}=\dfrac{259}{336}, so \angle CAD=39.571\ldots^\circ.
5. The required angle is made from one angle in each triangle.

6. \angle BAD=87.688\ldots^\circ+39.571\ldots^\circ=127.260\ldots^\circ, which is 127.3^\circ to 1 decimal place.

Find the part of the angle at C in each triangle, then use the sine rule.

1. In triangle ACD we know AC=14 and AD=17. To use the sine rule we need \angle ACD, which is part of \angle BCD=130^\circ.
2. Triangle ABC has three known sides. The side opposite \angle ACB is AB=13: \cos(\angle ACB)=\dfrac{14^2+10^2-13^2}{2\times14\times10}=\dfrac{127}{280}, so \angle ACB=63.026\ldots^\circ.
3. \angle ACD=130^\circ-63.026\ldots^\circ=66.973\ldots^\circ.
4. Split the interior angle at C, then use the derived angle in triangle ACD.

5. In triangle ACD, AD=17 is opposite \angle ACD and AC=14 is opposite \angle ADC: \sin D=\dfrac{14\sin66.973\ldots^\circ}{17}=0.75791\ldots
6. \angle ADC=\sin^{-1}(0.75791\ldots)=49.280\ldots^\circ. The obtuse option, 130.7^\circ, is impossible as 66.97^\circ+130.7^\circ>180^\circ. So \angle ADC=49.3^\circ to 1 decimal place.