E6.5 - Triangle area using sine

E6.5 - Triangle area using sine

Two sides and the angle between them can determine a triangle's area even when its height is hidden. You will derive the sine-area formula, preserve exact answers, and work backwards from an area while checking possible angle branches. This lesson is for Extended.

Area from two sides and the included angle

Can two triangles with the same two side lengths have the same area but very different shapes? Keep two sides fixed and compare an included angle of 40∘40^\circ with one of 140∘140^\circ. Their perpendicular heights are equal, even though one triangle leans outwards.

Take side bb as the base. The other known side aa makes included angle CC with it. Dropping a perpendicular gives height h=asin⁡Ch=a\sin C, so the familiar area 12bh\frac12 bh becomes 12absin⁡C\frac12 ab\sin C.

For obtuse CC, the foot lies on the extended base. The height is asin⁡(180∘−C)=asin⁡Ca\sin(180^\circ-C)=a\sin C, using the supplementary-angle relationship already established. The height remains positive.

[DIAGRAM: asset_name: The height behind the sine-area formula; asset_slug: area_height; caption: Supplementary included angles give equal perpendicular heights and equal areas.; description: Two exact triangles have adjacent sides a and b and included angles 40 and 140 degrees. In each the perpendicular height from the endpoint of a is a sine C. The obtuse triangle has its perpendicular foot on the extension of base b. Right-angle squares mark the feet, and equal heights are highlighted.]
Supplementary included angles give equal perpendicular heights and equal areas.

Area from two sides and their included angle

K=12absin⁡CK=\frac12 ab\sin C

KK is the area, a,ba,b are positive side lengths, and CC is the angle between them, with 0∘<C<180∘0^\circ<C<180^\circ. Use consistent length units; the area has their square unit.

Given in the List of formulas.

The letter choices can change, but “included” cannot. The required angle touches both chosen sides.

Worked example 1: an acute included angle

Question. A triangle has sides 8 cm8\,\mathrm{cm} and 11 cm11\,\mathrm{cm} enclosing an angle of 40∘40^\circ. Calculate its area.

Plan. Substitute the two enclosing sides and their included angle. Use degree mode.

K=12(8)(11)sin⁡40∘=44sin⁡40∘=28.2826…\begin{aligned} K&=\frac12(8)(11)\sin40^\circ\\ &=44\sin40^\circ\\ &=28.2826\ldots \end{aligned}

Answer. 28.3 cm228.3\,\mathrm{cm}^2, to 33 significant figures.

Check. Taking 11 cm11\,\mathrm{cm} as base gives perpendicular height 8sin⁡40∘=5.14230… cm8\sin40^\circ=5.14230\ldots\,\mathrm{cm}. Then 12(11)(5.14230…)=28.2826… cm2\frac12(11)(5.14230\ldots)=28.2826\ldots\,\mathrm{cm}^2.

Change only the included angle to its supplement. This deliberately tests whether the formula still works for an obtuse triangle.

Worked example 2: an obtuse included angle

Question. A triangle has sides 8 cm8\,\mathrm{cm} and 11 cm11\,\mathrm{cm} enclosing an angle of 140∘140^\circ. Calculate its area.

Plan. Use the same enclosing sides; sine handles the positive perpendicular height.

K=12(8)(11)sin⁡140∘=44sin⁡40∘supplementary angles have equal sine=28.2826…\begin{aligned} K&=\frac12(8)(11)\sin140^\circ\\ &=44\sin40^\circ && \text{supplementary angles have equal sine}\\ &=28.2826\ldots \end{aligned}

Answer. 28.3 cm228.3\,\mathrm{cm}^2.

Check. Reflecting the sloping side across the perpendicular leaves its height unchanged. The base is still 11 cm11\,\mathrm{cm}, so equal base and height must give equal area.

Watch out

If 88 and 1111 are the chosen sides, the formula needs the angle where those two sides meet. Substituting some other angle of the triangle may look tidy but does not give the correct perpendicular height.

Exact and numerical triangle areas

The same area formula can produce an exact surd or a decimal approximation. Decide which form the question requests before rounding anything.

You already know

sin⁡30∘=12\sin30^\circ=\frac12, sin⁡45∘=22\sin45^\circ=\frac{\sqrt2}{2}, and sin⁡60∘=32\sin60^\circ=\frac{\sqrt3}{2}. Substitute these values exactly on a non-calculator question.

Even on a calculator paper, retain exact values or full calculator precision during working. Round the final area once. If the angle is 90∘90^\circ, the formula becomes 12ab\frac12 ab, as it should for two perpendicular sides.

Worked example 3: an exact surd area

Question. A triangle has sides 10 cm10\,\mathrm{cm} and 14 cm14\,\mathrm{cm} enclosing 60∘60^\circ. Find its exact area.

Plan. Replace sine by its exact value and simplify.

K=12(10)(14)sin⁡60∘=70×32=353\begin{aligned} K&=\frac12(10)(14)\sin60^\circ\\ &=70\times\frac{\sqrt3}{2}\\ &=35\sqrt3 \end{aligned}

Answer. 353 cm235\sqrt3\,\mathrm{cm}^2.

Check. The height on base 1414 is 10(3/2)=5310(\sqrt3/2)=5\sqrt3, so 12(14)(53)=353\frac12(14)(5\sqrt3)=35\sqrt3. Also, 1<3<21<\sqrt3<2, so the area lies between 3535 and 70 cm270\,\mathrm{cm}^2.

Change only the angle to 65∘65^\circ. It no longer has one of the required exact special values, so a numerical answer is appropriate here.

Continue this lesson.

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