Solved Examples - Solution of Triangles | Mock Tests for JEE Main and Advanced 2026 PDF Download

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.