|The Law of sines|
In any triangle, the sides are proportional to the sines of the opposite angles, i.e.,
a/sin A = b/sin B = c.sin C
Proof: Let ABC be any oblique triangle. Draw the altitude CD = h. Two cases are possible; all angles are acute or one angle obtuse.
In either case;
Sin B = h/a from the first figure
h = a sin B
From the second figure
sin A = h/b
h = b sin A
Equating the two values of h we get, asin B = bsin A => a/sin A = b/sin B
Similarly by drawing perpendicular from A to BC, we can prove that
c/sin C = b/sin B
Therefore in any triangle a/sin A = b/sin B = c/sin C
Relations between three sides and two angles
In any triangle, to prove that
a = b cos C + c cos B, b = c cos A + a cos C, c = a cos B + b cos A
In the first figure we have
AD/AC = cos A , hence AD = AC cos A = b cos A
DB/BC = cos B, hence DB = BC cos B = a cos B
Now, AB = AD + DB = b cos A + a cos B => c = a cos B + b cos A
Similarly, we can prove the other relations.
The law of cosines
The square on any side of a triangle is equal to the sum of the squares on the other two sides minus twice the product of those two sides and the cosine of the included angle, i.e,
a2 = b2 + c2 - 2bc cos A
b2 = c2 + a2 - 2ca cos B
c2 = a2 + b2 - 2ab cos C
Proof: Consider any oblique triangle ABC with the altitude CD drawn from the vertex C to the opposite side.
From the first figure we have
AD = b cos A, CD = b sin A, DB = AB - AD = c-b cos A
From figure 2 we have
cos A = cos (180 - ang CAD) = -cos CAD = - AD/b => AD = -b cos A; CD = b sin A;
DB = DA + AB = c - b cos A
Thus in both figures, DB = c - b cos A, CD = b sin A
Hence a2 = DB2 + CD2
= (c-b cos A)2 + b2 sin 2A
=> a2 = c2 - 2bc cos A + b2(cos2A + b2Sin2A).
Hence a2 = b2 + c2 - 2bc cos A
Other results can be proved similarly.
Cos A = b2 + c2 - a2/2bc
Cos B = c2 + a2 - b2/2ca
Cos C = a2 + b2 - c2/2ab
Area of a triangle
A = s(s-a)(s-b)(s-c)1/2
s = (a+b+c)/2
where a, b and c are sides of the triangle .
Directions: Solve the following.