Properties and solutions of Triangles


Properties and solutions of Triangles is a vital component in the IIT JEE mathematics syllabus. With the help of practice, a person can get good hold on these topics which easily fetch 2-3 questions in IIT JEE. This head also comprises various formulae and rules like the sin rule, cosine rule, tangent rule etc. Sometimes questions are asked directly based on these results.

Various sub heads which come under this topic are enlisted below:

We list below some of the key points but the sub topics have been discussed in detail in the coming sections:



  • The cosine rule also holds true in any triangle: 
A triangle ABC with sides c, a, b opposite to angles C, A, B respectively

 
The cosine rule states that
c2= a2+b2-2ab cos(C)

It can be used to find the sides of a triangle.

Illustration 1: In a triangle ABC, 2ac sin 1/2 (A-B+C) = ?

Solution: In a triangle ABC,

A-B+C = 180° – 2B

Hence, 2ac sin [1/2 (A-B+C)] = 2ac sin (90°–B)

                                     = 2ac cos B

                                     = a2+c2-b2   (using the law of cosines)

Illustration 2: Which of the following pieces of data does not uniquely determine an acute angled triangle ABC (where R is the radius of the circumcircle)?

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

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

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

1. When a, sin A, sin B are given it becomes possible to find

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.

4. a, sin A, R is given, data is in sufficient 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 3: Comment on the statement whether it is possible to draw any tangent from the point (5/2, 1) to the circumcirle of the triangle with vertices (1, √3), (1, -√3), (3, √3).

Solution: Since, (1, √3), (1, -√3), (3, √3) form a right angle triangle at (1, √3)

Therefore, the equation of the circumcircle taking (3, √3) and (1, -√3) as end points of the diameter

So, (x-3) (x-1) + (y - √3) (y + √3) = 0.

This means, x2 - 4x +3 + y2 – 3 = 0.

Hence, x2 - 4x + y2 = 0.

At point (5/2, 1), S1 = 25/4 + 1 -10 < 0

Hence, point (5/2, 1) lies inside the circle.

Since, the point lies inside the circle, so it is not possible to draw any tangent.

For detailed knowledge of solutions of triangles, refer the video:

Students can also refer the papers of previous years to get an idea about the asked questions in IIT JEE Papers Provided by askIITians. 

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