HG23 - Area of any triangle

HG23 - Area of any triangle

Calculate a triangle's area from two sides and their included angle, even when its perpendicular height is unknown. Rearrange the same formula for a side or angle, and use a justified triangle split in larger shapes. This is Higher content.

The sine-area formula

Two sides of a triangle are 9 cm9\,\mathrm{cm} and 8 cm8\,\mathrm{cm}. Their included angle is 30∘30^\circ. How can you find the area without being given a height?

The included angle is the angle between the two sides used in the formula. Label BC=aBC=a, CA=bCA=b, and their included angle CC. If BCBC is the base, drop the perpendicular from AA to its line. For acute CC, sin⁡C=h/b\sin C=h/b, so the height is h=bsin⁡Ch=b\sin C.

Area=12×base×height=12a(bsin⁡C)=12absin⁡C.\text{Area}=\tfrac12\times\text{base}\times\text{height} =\tfrac12 a(b\sin C)=\tfrac12 ab\sin C.

[DIAGRAM: asset_name: Acute and obtuse triangle heights; asset_slug: sine-area-heights; caption: The perpendicular height is b sin C, including when its foot lies outside the triangle.; description: Two exact triangles with base BC 9 cm and side CA 8 cm, included angles C 30 and 150 degrees. Both heights are 4 cm; the obtuse triangle's altitude meets an extension of BC.]
The perpendicular height is b sin C, including when its foot lies outside the triangle.

For obtuse CC, the foot lies on the base line extended. The small right triangle uses 180∘−C180^\circ-C, giving h=bsin⁡(180∘−C)=bsin⁡Ch=b\sin(180^\circ-C)=b\sin C. Thus the same area formula works. At 90∘90^\circ, sine is 11, recovering half the product of the perpendicular sides.

Triangle area from an included angle

K=12absin⁡C.K=\tfrac12 ab\sin C.

KK denotes area; a,b>0a,b>0 are side lengths and 0∘<C<180∘0^\circ<C<180^\circ is their included angle. Use consistent length units; area has squared units. The formula is supplied on the current Higher additional formula sheet.

Choose the two sides enclosing the given angle, then substitute their lengths before evaluating the sine.

Worked example 1: an acute included angle

Question. Find the area when two sides are 9 cm9\,\mathrm{cm} and 8 cm8\,\mathrm{cm}, with included angle 30∘30^\circ.

Plan. Use these two sides and the angle between them.

K=12(9)(8)sin⁡30∘=36×12=18.K=\tfrac12(9)(8)\sin30^\circ =36\times\tfrac12=18.

Answer. 18 cm218\,\mathrm{cm^2}.

Check. The perpendicular height is 8sin⁡30∘=4 cm8\sin30^\circ=4\,\mathrm{cm}. Base–height area is 12(9)(4)=18 cm2\tfrac12(9)(4)=18\,\mathrm{cm^2}.

Keep the side lengths but open the included angle.

Worked example 2: an obtuse included angle

Question. Find the area when two sides are 9 cm9\,\mathrm{cm} and 8 cm8\,\mathrm{cm}, with included angle 150∘150^\circ.

Plan. Use the actual included angle, retaining its positive sine.

K=12(9)(8)sin⁡150∘=18.K=\tfrac12(9)(8)\sin150^\circ=18.

Answer. 18 cm218\,\mathrm{cm^2}.

Check. The external right triangle has angle 30∘30^\circ and height 4 cm4\,\mathrm{cm}. The same base and perpendicular height explain why the two differently shaped triangles have equal area.

For fixed sides, supplementary included angles have equal sine and hence equal area. The greatest possible area is 12ab\tfrac12 ab, when the included angle is 90∘90^\circ.

Watch out

The angle must sit between the chosen sides. A side opposite the given angle cannot simply be substituted as one of the two enclosing sides.

Use new lengths and a different angle.

Unknown measures

The area formula also determines an enclosing side when the area, other enclosing side and included angle are known. Divide by the entire coefficient of the unknown.

Rearranging the area formula

b=2Kasin⁡C,sin⁡C=2Kab.b=\frac{2K}{a\sin C}, \qquad \sin C=\frac{2K}{ab}.

For an angle, first evaluate 2K/(ab)2K/(ab), then apply inverse sine in degree mode and check the supplementary candidate. The letters must describe the same two sides and their included angle throughout.

For a nondegenerate triangle, the known side and the sine of its included angle are positive, so dividing by their product is valid.

Worked example 3: finding an enclosing side

Question. A triangle has area 30 cm230\,\mathrm{cm^2}, one side 8 cm8\,\mathrm{cm}, and included angle 30∘30^\circ between that side and side bb. Find bb.

Plan. Write the area equation and isolate bb.

30=12(8)bsin⁡30∘=2b,b=15.30=\tfrac12(8)b\sin30^\circ=2b, \qquad b=15.

Answer. b=15 cmb=15\,\mathrm{cm}.

Check. Using bb as the sloping side gives height 15sin⁡30∘=7.5 cm15\sin30^\circ=7.5\,\mathrm{cm}. Then 12(8)(7.5)=30 cm2\tfrac12(8)(7.5)=30\,\mathrm{cm^2}.

An unknown angle can have two candidates because supplementary angles have the same sine.

Worked example 4: selecting an obtuse angle

Question. A triangle has sides 6 cm6\,\mathrm{cm} and 10 cm10\,\mathrm{cm}, area 15 cm215\,\mathrm{cm^2}, and an obtuse included angle CC. Find CC.

Plan. Isolate sine, then use the stated obtuse condition.

sin⁡C=2(15)6(10)=12.\sin C=\frac{2(15)}{6(10)}=\frac12.

Inverse sine gives 30∘30^\circ. The other candidate is 180∘−30∘=150∘180^\circ-30^\circ=150^\circ.

Answer. C=150∘C=150^\circ.

Check. With base 6 cm6\,\mathrm{cm}, the altitude must be 2K/6=5 cm2K/6=5\,\mathrm{cm}. A 10 cm10\,\mathrm{cm} sloping side reaches that height with the 30∘30^\circ external angle, consistent with the 150∘150^\circ interior angle.

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