A+ » VCE » Further Maths U3 & 4 Master Notes » OA3 Geometry and Measurement » 1.6 Solving Triangles in 2 and 3 dimensions

1.6 Solving Triangles in 2 and 3 dimensions

Note: if you cannot remember the trigonometric identities, revise notes for 1.5 Methods for Solving Triangles.

Solving for Area without Height

  • In some cases, triangles will be given with the edge lengths rather than the height. In this case our regular formula for the area of a triangle (A=\frac{1}{2} b h) needs some reworking. If we draw a line down the middle of a triangle, we can effectively split it into two right-angled triangles from which we can use our trigonometric identities to solve for the height. Using our formula for sine, we find:

\sin (A) =\frac{h}{c}

h =c \sin (\theta)


A=\frac{1}{2} b c \sin (\theta)

Picture 8
  • This formula works for non-right angled triangles as well.

Note: trigonometry is explored in more detail in notes 2.3 Solving Triangles using Trigonometry and 2.4 Applications of Trigonometry and Pythagoras Theorem.

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