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```Solved Examples on Solution of Triangles

Illustration 1: Which of the following pieces of data does not uniquely determine an acute angled triangle ABC (where R is the radius of the circumcircle)? (2002)

1. a, sin A, sin B                                                  2. a, b, c

3. a, sin B, R                                                       4. a, sin A, R

Solution: We shall discuss all the four parts one by one.

1. When a, sin A, sin B are given it becomes possible to find the rest of the sides as

b = a sin B/ sin A,

c = a sin C/ sin A

So, all the three sides are unique.

2. The three sides can uniquely make an acute angled triangle.

3. a, sin B, R is given and so the data is sufficient for computing the remaining sides and triangles.

b = 2R sin B, sin A = a sin B/b

So, sin C can be determined. Hence, this also implies that the side c can be uniquely determined.

4. a, sin A, R is given, data is insufficient to find the other sides and angles

b/ sin B = c/ sin C = 2R

Hence, it is clear that this could not determine the exact values of a and b. Illustration 2:  Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose, a = 6, b = 10 and the area of the triangle is 15√3. If ACB is obtuse and if r denotes the radius of the incircle of the triangle, then find the value of r2. (2010)

Solution: The two sides ‘a’ and ‘b’ of the triangle are given to be 6 and 10 respectively.

Now, sin C = √3/2 and C is given to be obtuse

This gives C = 2π/3 = √(a2 + b2 - 2ab cos C)

= √62 + 102 – 2.6.10. cos 2π /3

= 14

Hence, r = ?/s which means

Illustration 3: If ? is the area of triangle with sides of length a, b and c then show that ? ≤ 1/4 √ (a+b+c) abc. Also show that the equality occurs in the above inequality if and only if a = b = c. (2001)

Solution: The area of the triangle is ? and the lengths of the sides are a, b and c.

Now, it is given that ? ≤ 1/4 √(a+b+c) abc

Hence, 1/4?. √(a+b+c) abc ≥ 1

So, (a+b+c) abc /16?2 ≥ 1

Hence, 2sabc/ 16?2 ≥ 1

So, sabc/ 8s(s-a)(s-b)(s-c) ≥ 1

abc/ 8(s-a)(s-b)(s-c) ≥ 1

Hence, abc/ 8 ≥ (s-a)(s-b)(s-c)

Now, put (s-a) = x ≥ 0, (s-b) = y ≥ 0, (s-c) = z ≥ 0

Hence, (s-a) + (s-b) = x + y

So, this gives 2x – a – b = x + y

So, c = x + y

Similarly, as a = y + z, b = x + z, so we get,

(x+y)/2. (y+z)/2. (x+z)/2 ≥ xyz which is true

Now equality will hold only if x = y = z i.e. only if a = b = c.

So, the equality will hold only if the triangle is equilateral.

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