 Hello and welcome to the session. The given question says, using vectors, proof that in triangle ABC, A divided by sine A is equal to small b divided by sine capital B is equal to small c divided by sine capital C, where small a, b and c are the lengths of sides opposite respectively to the angles a, b and c of triangle ABC. Let's now start with the solution and let vectors a, b and c denote the vectors represented at the sides respectively of triangle ABC as shown in this figure. So this implies that sum of vectors a, b and c is equal to zero vector, let this be equation number one. Now taking cross product of vector a on both the side, on the left hand side we have vector a cross, the sum of vectors a, b and c and on the right hand side we have vector a cross zero vector. This further implies a vector cross a vector plus a vector cross b vector plus a vector cross c vector is equal to zero vector and cross product of vector a with vector a is zero. So this further implies that we have cross product of vector a and b is equal to minus of cross product of vector a and c which further implies that a vector cross b vector is equal to c vector cross a vector. Let this be equation number two. Similarly cross multiplying one by vector b we get vector cross a vector plus b vector cross c vector is equal to zero vector which further implies cross product of vector b and c is equal to minus of cross product of vector b and a which is equal to cross product of vector a and b. Let this be equation number three. Then from two and three we have vector a cross vector b is equal to vector b cross vector c is equal to vector c cross vector a. Therefore their magnitudes will also be equal that is magnitude of cross product of vector a and b is equal to magnitude of cross product of vector b and c is equal to magnitude of cross product of vector c and a. This further implies that magnitude of vector a into magnitude of vector b into the sign of angle between the vectors a and b and the angle between the vectors a and b is pi minus c. So here we have pi minus c is equal to magnitude of vector b into magnitude of vector c into sign of angle between the vectors b and c which is pi minus a and this is further equal to magnitude of vector c into magnitude of vector a into sign of angle between c and a which is pi minus b. Now dividing throughout the magnitudes of vector a b reciprocals we get first let us divide so on dividing we have sign pi minus c divided by magnitude of vector c is equal to sign pi minus a divided by magnitude of vector a is equal to sign pi minus b divided by magnitude of vector b. And now let us take reciprocals magnitude of vector a divided by sign since I sign pi minus c times sign theta this is further equal to magnitude of vector b divided by sign b is equal to magnitude of vector c divided by sign c. And since vector b c is equal to a vector therefore magnitude of vector b c is equal to a so here we have a divided by sign a is equal to b divided by sign b is equal to c divided by sign c. This is also called the law of signs for a triangle so this is what we are required to prove this completes the fashion buy and take care.