Right triangle given one angle and hypotenuse (HA)

This page shows how to construct a right triangle that has the hypotenuse (H) and one angle (A) given. It works in three steps:

  1. Copy the angle A. (See Copying an angle)
  2. Copy the length of the hypotenuse onto the angle leg (See Copying a segment)
  3. Drop a perpendicular from the end of the hypotenuse. (See Perpendicular to a line from a point)

Printable step-by-step instructions

The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.


  Argument Reason
We first prove that ∆BCA is a right triangle
1 m∠BCA = 90° BC was constructed using the procedure in Perpendicular to a line from a point. See that page for proof.
2 Therefore ∆BCA is a right triangle By definition of a right triangle, one angle must be 90°
Now prove BA is the hypotenuse H
3 AB = the given hypotenuse H AB was copied from H at the same compass width
Now prove ∠BAC is the given angle A
4 m∠BAC = given m∠A Copied using the procedure in Copying an angle. See that page for proof
9 ∆BCA is a right triangle with the desired hypotenuse H and angle A From (2), (3), (4)

  - Q.E.D

Try it yourself

Click here for a printable worksheet containing two HA triangle construction problems. When you get to the page, use the browser print command to print as many as you wish. The printed output is not copyright.

Other constructions pages on this site




Right triangles

Triangle Centers

Circles, Arcs and Ellipses


Non-Euclidean constructions