Practical_Antenna_Handbook_0071639586

A p p e n d i x A : U s e f u l M a t h 721
30˚
d 2 = h
2 + (
d )
2
2
d
h
60˚
d
2
d
2
Figure A.4.5 Height of an equilateral triangle.
rior angles of each is 90 degrees, and we can see (Fig. A.4.6) that, from symmetry, each
of the other two interior angles of either isosceles triangle must be 45 degrees. We can
find the length of the shared side of the two triangles by recognizing that a = b and proceeding
as follows:
2 2 2 2 2
h = a + b or h = 2a
(A.4.9)
45˚
h
45˚
a
h
h = a 2
45˚
45˚
h
2
=
a
2
b = a
Figure A.4.6 Bisecting a square.