Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**narvel****Member**- Registered: 2014-04-06
- Posts: 2

If I have an isosceles triangle and know everything except the length of the base, is there a way to solve it other than using the law of sines or cosines? In other words, I know all the angles and the lengths of the two equal sides, and want to know if there is a way to find the length of the other side using a method other than the law of sines or cosines.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Wiki wrote:

A triangle with exactly two equal sides has exactly one axis of symmetry, which goes through the vertex angle and also goes through the midpoint of the base. Thus the axis of symmetry coincides with (1) the angle bisector of the vertex angle, (2) the median drawn to the base, (3) the altitude drawn from the vertex angle, and (4) the perpendicular bisector of the base.[4]

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**narvel****Member**- Registered: 2014-04-06
- Posts: 2

Bobbym, I suspect that your quote is a hint to use the height of the triangle to create two right triangles, and then use the Pythagorean Theorem. Thats kind of what Im looking for, an algebraic solution instead of a trigonometric solution, but I dont know how to find the height of the triangle without using trigonometry.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You will have to hunt for this yourself but I think the perpendicular bisector of the base passes through the vertex.

I was thinking of using SOHCAHTOA now, not Pythagoras.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,322

hi narvel,

Welcome to the forum.

The laws of sines and cosines are useful in triangles with no right angle. All isosceles triangles can be split into two equal triangles along the line of symmetry and that creates a right angled triangle.

In my diagram BD = AB . cos(ABC) and then you can double this to get BC.

So you would be using 'simple' trig. rather than the sine rule.

There are many triangles that could be drawn so that AB = AC = some known measurement. The angles fix a single triangle so you won't find a method that avoids using the angles completely.

Hope that's what you wanted,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline