# The Plemelj Construction of a Triangle: 8

The Plemelj Construction of a Triangle: 8

This Demonstration constructs a triangle given the length of its base, the length of the altitude from to and the difference between the angles at and . This is not one of Plemelj's original constructions, but a new one based on his equation , where and .

ABC

AB=c

h

C

C

AB

α-β

A

B

msin(γ-μ)=ccos(α-β)

m=+

2

c

2

(2)

h

C

μ=arctan(λ/2)

h

C

The modified equation is , which can be read as the law of sines for a triangle where the sides of length 1 and are opposite the angles and , respectively, or, multiplying by , that the sides of length and are opposite the angles and .

sin(γ-μ)=sin(μ)cos(α-β)

cos(α-β)

μ

γ-μ

c

c

ccos(α-β)

μ

γ-μ

Construction

Step 1: Draw a straight line of length and a line parallel to at distance . Let be the midpoint of and the point on just above .

AB

c

τ

AB

h

C

M

AB

N

τ

M

Step 2: Draw a circle with center and radius .

σ

M

c/2

Step 3: Draw the ray from at the angle δ from to intersect at the point .

A

AB

σ

P

Step 4: Draw the circle with center and radius .

ζ

N

m/2

Step 5: Measure out a point on the circle at distance from .

D

ζ

AP=ccosδ

A

Step 6: The point is the intersection of and the right bisector of .

C

τ

DP

Step 7: The triangle meets the stated conditions.

ABC

Verification

According to the law of sines, the exterior angle at of is , which is . The angle is . So and .

A

ΔADB

γ

∠ADB+∠ABD=μ+γ-μ

DAC

(π-γ-δ)/2=α-δ

(α+β)-δ=2α-2δ

δ=α-β