Every triangle has 3 angles and 3 sides. Given any 3 quantities out of 3 angles and 3 sides (at least one of which is a side) are given then remaining 3 can be found , which is termed as solution of the triangle. Many interesting relation among these quantities will be discussed in this section. Which formula is to be used where is a very important point here, as a use of wrong formula might lead to large calculations. So be prepared to see some very interesting stuff.
Side AB , BC, CA of a DABC are denoted by c, a, b respectively and is semi perimeter of the triangle and D denotes area of the triangle.
Draw AD perpendicular to opposite side meeting it in point D.
Þ AD = cSinB
In D ACD,
Þ AD = bSinC
Equating 2 values of AD we get
cSinB = bSinc
Similarly drawing a perpendicular line from B upon CA we have
Given that ÐB=30, C=10 and b=5 find the angles of A and C of triangle.
By using sine rule,
Let ABC be a triangle and let perpendicular from A on BC meet it in point D
Now AB2= AD2+BD2
= (BC-CD) 2+ (AC2-DC2)
= AC2+BC2-2BC.CD ------------------ (1)
So, equation (1) is now
c2 = b2+a2-2abCosC
Similarly CosA, CosB can be found.