 Hello and welcome to the session. In this session we will discuss the variation of law of cosines and we will use them to solve triangles. Now let us discuss law of cosines. Now let triangle ABC be any triangle with small a, small b and small c representing the variations of sides opposite to angles with measurements a, b and c respectively. Then first is a square is equal to b square plus c square minus 2 BC cos of capital A. Second is b square is equal to a square plus c square minus 2 AC into cos of capital B. Third is c square is equal to a square plus b square minus 2 AC into cos of capital C. Firstly we will prove a square is equal to b square plus c square minus 2 BC into cos of capital A. For this we develop a relationship between sides a, b and c and angle a from the given triangle. Now let us draw a perpendicular from the vertex b meeting the side AC at point b. So bd is the height of triangle ABC given by small h. Now let length of side AD be x and length of side AC is b. So length of side DC will be b minus x. Now triangle BDC is the right angle triangle where BD is equal to h, BC is equal to b minus x and BC is equal to small a. So in triangle BDC by a perpendicular form we have BD square plus BC square is equal to BC square. Now let us put the values of these sides and we have x square plus b minus x whole square is equal to a square. This implies x square plus b minus x whole square will be b square plus x square minus 2 Bx is equal to a square. Now here triangle ABT is again a right angle triangle. So in triangle ABD by Pythagoras theorem we have BG square plus AB square is equal to AB square. Now putting the values of sides we have x square plus x square is equal to c square. Now let this be equation number 1 and this be equation number 2. Now from right angle triangle ABD we can find the value of cos of angle A which will be equal to adjacent side upon height of news. For triangle A the adjacent side will be side AD and height of news is AB. So cos of angle A is equal to x upon c. This implies x is equal to c into cos of angle A. Let this be equation number 3. Now using equation number 3 and equation number 2 in equation 1 we eliminate h and x from equation number 1. So we have from equation 1 x square plus b square plus x square minus 2 Bx is equal to a square which can be written as x square plus x square plus b square minus 2 Bx is equal to a square. Now here from equation number 2 we can see x square plus x square is equal to c square. So here it will be c square plus b square minus 2 B into x. Now from equation 3 we have x is equal to c into cos of A. So here it will be minus 2B into c into cos of A which is equal to a square. So this implies a square is equal to b square plus c square minus 2 BC into cos of angle A. So we have derived this first equation. Similarly we can derive these two equations also by drawing the perpendicular from vertices c respectively to the opposite sides. Now let us solve triangles using law of cosines. Now law of cosines are used to solve a triangle when we are given side angle side that is measure of two sides and their included angle and secondly side side side that is measure of all the three sides of the triangle. Now when side angle and side is given then let us see how to use law of cosines to solve the triangle. Now let us consider the following example. Here we can see two sides and their included angle is given to us and we have to find all missing dimensions of triangle ABC. So we have to find angles c also we have to find side given by small b. Now when we are given angle b so we will use the second equation of law of cosines and we can find length of side b. So putting the values of sides A and c and also angle b we have b square is equal to 13 square plus 20 square minus 2 into 13 into 20 into cos of angle b that is cos of 102 degrees this implies b square is equal to 169 plus 400 minus 520 into now cos of 102 degrees is minus 0.207 and this implies b square is equal to 569 plus 107.64 this implies b square is equal to 676.64. Now taking positive square root we get b is approximately equal to 26. Now to find the angles we can use law of sines that is sin of angle p upon b is equal to sin of angle c upon c. Now putting the values we have sin of angle 102 degrees upon 26 is equal to sin of angle c upon 20 which implies 0.98 upon 26 is equal to sin c upon 20 Also in this we have sin of angle c is equal to 0.754 and this implies angle c is equal to sin inverse of 0.754 which implies angle c is approximately equal to 48.9 degrees. Now here we have used law of sines to find angle c or we can use this equation of law of sines to find angle c. Now putting the values here we have 20 square is equal to 13 square plus 26 square minus 2 into 13 into 26 into cos of angle c and on solving this we get cos of angle c is equal to 0.658 and this implies angle c is equal to cos inverse of 0.658 which implies angle c is approximately equal to 48.3 degrees. Now using law of sines we have got angle c is equal to 48.9 degrees and the law of cosines we have got angle c as 48.3 degrees so both laws do same measure of angle c that is 38 degrees approximately. Now we have to find angle a. Now employ angle apc by using some angle property angle a plus angle b plus angle c is equal to 180 degrees which implies angle a plus 102 degrees plus 48.9 degrees is equal to 180 degrees and on solving this we get angle a is equal to 29.1 degrees. Now note that in this example both the angles a and c are acute angles so we could have found even angle a using law of sines as we have found angle c but in some cases one of the angles to be found might be obtuse so always make drawing of triangle to know whether the angle will be obtuse or not. Never use law of sines to find the obtuse angle because law of sines will never produce an obtuse angle so use law of sines to find a acute angle. Now let us see how we can use law of cosines to find all the measurements of the given triangle when all three sides of this triangle are given. Now when we are given the sides a b and c and we have to find measure of all the three angles a b and c we first use any one of the law of cosines to find one of the angles. Once we have an angle then we can either use another law of cosine equation or law of sines to find another angle now the law of sines will never produce an obtuse angle. If an angle is obtuse never use the law of sines to find it in this case make use of law of cosines to find the largest possible angle first. So in this session we have proved law of cosines and we have learnt how to use them to solve triangles and this completes our session hope you all have enjoyed the session.