Area of a Triangle Using Sine

The angle must be between the two sides in the formula.

1. Area =\dfrac12ab\sin C, where C is the angle between the sides a and b.
2. Here 36^\circ is between AB=8 and AC=11, so area =\dfrac12\times8\times11\times\sin36^\circ.
3. Without the \dfrac12 you get double the area; \cos is the wrong ratio; and \dfrac12\times8\times11 only works when the angle is 90^\circ.

Use the angle between the two given sides.

1. 47^\circ is between the two given sides, so use area =\dfrac12ab\sin C.
2. Area =\dfrac12\times9\times6\times\sin47^\circ=19.746\ldots
3. The area is 19.7 cm^2 to 1 decimal place.

Use the angle between the two given sides.

1. 128^\circ is between the two given sides, so use area =\dfrac12ab\sin C. The formula works for obtuse angles too.
2. Area =\dfrac12\times12\times7\times\sin128^\circ=33.096\ldots
3. The area is 33.1 cm^2 to 1 decimal place.

Use the exact value of the sine of a right angle.

1. The 90^\circ angle is between the given sides, so area =\dfrac12\times8\times6\times\sin90^\circ.
2. \sin90^\circ=1, so the area is \dfrac12\times8\times6=24 cm^2. This matches \dfrac12\times\text{base}\times\text{height} for a right-angled triangle.

Locate the vertex where the two stated sides meet.

1. In area =\dfrac12ab\sin C, the angle must be the one between the two sides used.
2. AB and BC meet at B, so \theta=\angle ABC=54^\circ.
3. 83^\circ is at A and 43^\circ is at C, so neither is the angle between AB and BC. 90^\circ would only be right if the triangle had a right angle at B, and it does not.

Substitute into the area formula and isolate the missing side.

1. 48^\circ is between AB and AC, so area =\dfrac12\times AB\times AC\times\sin48^\circ.
2. 32.4=\dfrac12\times AB\times9\times\sin48^\circ. Multiply both sides by 2: 64.8=AB\times9\sin48^\circ.
3. 9\sin48^\circ=6.6883\ldots, so AB=\dfrac{64.8}{6.6883\ldots}=9.688\ldots
4. AB=9.7 cm to 1 decimal place.

First find the sine of the angle from the area formula.

1. AB=8 and AC=10 enclose \angle BAC, so 24=\dfrac12\times8\times10\times\sin A.
2. 24=40\sin A, so \sin A=\dfrac{24}{40}=0.6.
3. \angle BAC=\sin^{-1}(0.6)=36.869\ldots^\circ, which is acute as required.
4. \angle BAC=36.9^\circ to 1 decimal place.

Inverse sine gives an acute angle, but the obtuse angle 180^\circ-x has the same sine.

1. AB and AC enclose \angle BAC: 35=\dfrac12\times10\times9\times\sin A=45\sin A, so \sin A=\dfrac{35}{45}=\dfrac79.
2. \sin^{-1}\left(\dfrac79\right)=51.057\ldots^\circ, but this is acute.
3. Since \sin(180^\circ-x)=\sin x, the obtuse angle with the same sine is 180^\circ-51.057\ldots^\circ=128.942\ldots^\circ.
4. \angle BAC=128.9^\circ to 1 decimal place.

Convert the side lengths to the same unit before using the area formula.

1. The answer is in square metres, so change both sides to metres: 160 cm =1.6 m.
2. The 64^\circ angle is between the sides, so area =\dfrac12\times1.6\times2.4\times\sin64^\circ=1.7256\ldots
3. The area is 1.7 m^2 to 1 decimal place.

A diagonal splits a parallelogram into two triangles of equal area.

1. The diagonal BD splits the parallelogram into two congruent triangles, ABD and CDB, with equal areas.
2. Triangle ABD: 52^\circ is between AB=11 and AD=7, so its area is \dfrac12\times11\times7\times\sin52^\circ=30.338\ldots cm^2.
3. Parallelogram area =2\times30.338\ldots=60.676\ldots, which is 60.7 cm^2 to 1 decimal place.

Find the two triangle areas separately, using the diagonal as one side in each.

1. Split ABCD along AC into triangles ABC and ACD, and use \dfrac12ab\sin C in each.
2. Triangle ABC: 35^\circ is between AC=12 and AB=8, so area =\dfrac12\times12\times8\times\sin35^\circ=27.531\ldots cm^2.
3. Triangle ACD: 54^\circ is between AC=12 and CD=7, so area =\dfrac12\times12\times7\times\sin54^\circ=33.978\ldots cm^2.
4. Total area =27.531\ldots+33.978\ldots=61.510\ldots, which is 61.5 cm^2 to 1 decimal place.

Use the ratio to find the part of the base used by the smaller triangle.

1. AD:DB=2:3, so AD is \dfrac25 of AB: AD=\dfrac25\times15=6 cm.
2. Triangle ACD has sides AD=6 and AC=10 enclosing \angle CAD=47^\circ.
3. Area =\dfrac12\times6\times10\times\sin47^\circ=21.940\ldots, which is 21.9 cm^2 to 1 decimal place.

Write an area expression for each triangle and equate them.

1. Use the angle between the two sides in each triangle: area of ABC=\dfrac12\times10\times7\sin38^\circ and area of ABD=\dfrac12\times10\times9\sin(\angle BAD).
2. The areas are equal: \dfrac12\times10\times9\sin(\angle BAD)=\dfrac12\times10\times7\sin38^\circ.
3. Divide both sides by \dfrac12\times10=5: 9\sin(\angle BAD)=7\sin38^\circ, so \sin(\angle BAD)=\dfrac{7\sin38^\circ}{9}=0.47884\ldots
4. \angle BAD=\sin^{-1}(0.47884\ldots)=28.610\ldots^\circ (acute), which is 28.6^\circ to 1 decimal place.

Find the perpendicular height from the sloping side.

1. Area of a trapezium =\dfrac12(a+b)h, where h is the perpendicular height between the parallel sides.
2. The perpendicular from D to AB makes a right-angled triangle with hypotenuse AD=9 and angle 64^\circ at A, so h=9\sin64^\circ=8.0891\ldots cm.
3. Area =\dfrac12\times(14+8)\times8.0891\ldots=88.980\ldots
4. The area is 89.0 cm^2 to 1 decimal place.

Use the area formula to form an equation in the square of the side length.

1. The angle 50^\circ is between the sides x and 2x, so 36=\dfrac12\times x\times2x\times\sin50^\circ.
2. \dfrac12\times x\times2x=x^2, so x^2\sin50^\circ=36.
3. x^2=\dfrac{36}{\sin50^\circ}=46.994\ldots
4. x=\sqrt{46.994\ldots}=6.855\ldots (a length is positive), so x=6.9 to 1 decimal place.

Use the common base to simplify the ratio of the two area expressions.

1. Both triangles have the side AB=12 and use the angle at A: area of ABC=\dfrac12\times12\times9\sin36^\circ and area of ABD=\dfrac12\times12\times AD\sin74^\circ.
2. The ratio of the areas is 3:5, so \dfrac{\tfrac12\times12\times9\sin36^\circ}{\tfrac12\times12\times AD\sin74^\circ}=\dfrac35. The \dfrac12\times12 cancels: \dfrac{9\sin36^\circ}{AD\sin74^\circ}=\dfrac35.
3. Cross-multiply: 3\times AD\sin74^\circ=5\times9\sin36^\circ, so AD=\dfrac{45\sin36^\circ}{3\sin74^\circ}.
4. AD=9.172\ldots, which is 9.2 cm to 1 decimal place.

Subtract the known triangle area before recovering the unknown angle.

1. The diagonal AC splits ABCD into triangles ABC and ACD.
2. Triangle ABC: 42^\circ is between AC=11 and AB=7, so its area is \dfrac12\times11\times7\times\sin42^\circ=25.761\ldots cm^2.
3. So triangle ACD has area 70-25.761\ldots=44.238\ldots cm^2.
4. In triangle ACD, AC=11 and AD=9 enclose \angle CAD: 44.238\ldots=\dfrac12\times11\times9\times\sin(\angle CAD), so \sin(\angle CAD)=\dfrac{2\times44.238\ldots}{99}=0.89370\ldots
5. \angle CAD=\sin^{-1}(0.89370\ldots)=63.342\ldots^\circ (acute), which is 63.3^\circ to 1 decimal place.

Subtract the triangle area from the sector area.

1. Segment area = sector area - area of triangle AOB.
2. Sector area =\dfrac{100}{360}\times\pi\times10^2=87.266\ldots cm^2.
3. Triangle AOB has two radii of 10 cm enclosing 100^\circ: area =\dfrac12\times10\times10\times\sin100^\circ=49.240\ldots cm^2.
4. Segment area =87.266\ldots-49.240\ldots=38.026\ldots, which is 38.0 cm^2 to 1 decimal place.

Find the angle between the known sides before using the area formula.

1. The given angle \angle ABC=118^\circ is not between AB and AC; the angle between them is \angle BAC. Find it using the sine rule.
2. AC=15 is opposite \angle B, and AB=9 is opposite \angle C: \sin C=\dfrac{9\sin118^\circ}{15}=0.52976\ldots, so \angle C=31.989\ldots^\circ. It must be acute, because \angle B is already obtuse.
3. Angles in a triangle: \angle BAC=180^\circ-118^\circ-31.989\ldots^\circ=30.010\ldots^\circ.
4. Area =\dfrac12\times9\times15\times\sin30.010\ldots^\circ=33.760\ldots
5. The area is 33.8 cm^2 to 1 decimal place.

Find an angle from the three sides, then use the sine area formula.

1. No angle is known, so use the cosine rule to find one, then area =\dfrac12ab\sin C. In the diagram, \angle A is between the sides 10 m and 13 m, and the side opposite it is 17 m.
2. \cos A=\dfrac{10^2+13^2-17^2}{2\times10\times13}=\dfrac{-20}{260}=-0.07692\ldots, so \angle A=94.411\ldots^\circ.
3. Area =\dfrac12\times10\times13\times\sin94.411\ldots^\circ=64.807\ldots
4. The area is 64.8 m^2 to 1 decimal place.