Expert: Josh - 11/9/2004

Question
i have an isosceles triangle and have all the lenths, how do i find the angles?

Answer
Let's draw a diagram.

A
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D--------C
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B
An isosceles triangle is formed by joining ABC.
The dimensions |AB|,|BC| and |CA| are all known.

From the diagram, it is clear that |AD|=|DB| and |AC|=|CB|.
If we let |AD|=a, then, |AB|=2a.
Also, let |AC|=b, say. (Remember, the question tells us exactly what "a","b" represent. In particular, "a" is half the length of side AB)

Our aim is to work out angles ^ACB,^CBD and ^BAC.
Given: Angle ^ADC = ^BDC = 90 degrees, since CD is the perpendicular bisector of side AB.
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Relating angles to ratios between two sides of the triangle

Recall the following definitions with respect to the angle:
sine   = opposite_side/hypotenuse
cosine = adjacent_side/hypotenuse

*hypotenuse is the longest side of a right angle triangle.

So, sin(^CAD)=|DC|/|AC|=sqrt(b^2-a^2)/b ...using Pythagoras [#1]
   cos(^CAD)=|DA|/|AC|=a/b ...[#2]
Taking the inverse of cosine, [#2] gives the angle
^CAD = cos^(-1)(a/b)

We can work out the angle ^ACB in two ways.

Method 1:
^CBD = ^CAD = cos^(-1)(a/b),
angle sum of triangle adds up to 180 degrees,
so, ^ACB = 180 - (^CBD + ^CAD) = 180 - 2*cos^(-1)(a/b)

Method 2:
^ACD = 180-(^CDA + ^CAD)
    = 180-(90+cos^(-1)(a/b))
    = 90 - cos^(-1)(a/b)
^ACB = ^ACD + ^DCB
    = 2* (^ACD)
    = 180 - 2*cos^(-1)(a/b)

Any question, let me know.

Cheers.