6.3 Special Right Triangles
The 30^{0}  60^{0}  90^{0}
triangle : If the angles of a triangle are 30^{0}
, 60^{0} and 90^{0} then the side
opposite to 30^{0} is half the hypotenuse and the
side opposite to 60^{0} is
times the hypotenuse.
Figure 6.5
In D ABC Ð
A = 90^{0} Ð B = 60^{0}
and Ð C = 30^{0}
To prove that l
(AB) =
l (BC) and l
(AC) =
Take a point D on ray BA such that seg.AD @
seg.AB and join CD.
In D ABC and D
ACD
AB @ AD construction
Ð CAB @ Ð
CAD both are right angles
AC @ AC common side
\ D BAC @ D
DAC (SAS)
\ Ð ACB @
Ð ACD corresponding angles of congruent triangles are
equal.
but m Ð ACB = 30^{0}
\ m Ð
DCB = 60^{0}
Þ D DCB is an equilateral triangle
\ l
(seg.DC) = l (seg.BC) = l
(seg.DB) ® (1)
but l (seg. AB) =
l (DB) ® (2)
®(3) from (1) and (2).
In right triangle ABC
l (seg.AB)^{2} + l
(seg.AC)^{2} = l (seg.BC)^{2} Pythagoras
{
l (seg.BC)^{2} } + l
(seg.AC)^{2} = l (seg.BC)^{2}from
(3)
l (seg.AC)^{2} = l
(seg.BC)^{2} 
l (seg.BC)^{2}
=
l (BC)^{2}
The 45^{0}  45^{0}  90^{0} triangle
: If the angles of a triangle are 45^{0}  45^{0}
 90^{0} then the perpendicular sides are
times the hypotenuse.
In D ABC Ð
A = 45^{0} , Ð B = 90^{0}
and Ð C = 45^{0}.
Figure 6.6
To prove that AB = BC =
By Pythagoras theorem
l (seg.AB)^{2} + l
(seg.BC)^{2} = l (seg.AC)^{2}
l (seg.AB) = l
(seg.BC) . D ABC is isosceles.
\ l
(seg.AB)^{2} + l (seg.BC)^{2}
= 2 l (seg.AB)^{2} =
l (seg.AC)^{2}
