cosine law

Hello friends, In the previous post we have discussed about Derivative of cot and today we are going to discuss a very important topic cosine law, as we are now familiar with trigonometric parameters. All the parameters have their specific role in trigonometry and we use all the parameters according to our need. As we well know that the six parameters are sin, cos, tan, sec, cot and cosec. As we well know that we can apply all these law to a triangle and we also know that any triangle consists of three angle and three sides, now if we talk about cosine law then  if we are having the value of  two sides of a triangle and we are also having the value of the angle opposite to the side of triangle which we want to measure, suppose if a triangle consists of three sides as a ,b and c and if the value of  b and c are given and we are asked to find side a by cosine law then we must have the value of angle A.  We can write cosine law as,

a2 = b2 + c2 – 2bccosA

Here a, b and c are sides of the triangle and A is the angle between them.

Now we will see an example, in which we will apply cosine law

Example:  A triangle is having sides as b= 2 cm and c = 3cm and angle A = 60 degree find the side a by cosine law?

Solution: as we know by cosine law,

a2 = b2 + c2 – 2bccosA

Now we will put our values in this,

a2 = 4 + 6 – 2 *2*3 *1/2

a2 = 10 -6

a2 = 4

a = 2

This is the required solution of the given problem by cosine law.

 If you are having cbse class 10 sample papers then go through First Derivative Test, it is an important topic in mathematics.