 Hi and welcome to the session. Today we will learn about trigonometric ratios. Suppose we are given a right angle triangle ABC, right angle at B and we need to find out the trigonometric ratios with respect to angle A. So we know for angle A the side opposite to angle A is the perpendicular and we will denote the perpendicular by the letter P. The side adjacent to angle A is the base which will be denoted by B. AC is the height continuous as this is the side opposite to the right angle that is angle B. So it will be denoted by H. So now let us define the trigonometric ratios for angle A. First of all sin A is equal to P by H that is perpendicular by hypotenuse which will be equal to BC upon AC. Now cos A is equal to V by H that is base upon hypotenuse so it will be AB upon AC. Now tan A is equal to P upon B that is perpendicular upon base so this will be BC upon AB. Now cosecant A is 1 upon sin A so this will be H upon P that is hypotenuse upon perpendicular that will be AC upon BC. Also secant A is 1 upon cose so this will be H upon B that is hypotenuse upon base which will be AC upon AB. And lastly cot A is 1 upon tan A so this will be B upon P that is base upon perpendicular so this will be AB upon BC. Now in triangle ABC angle B is 90 degrees that means sum of angle A and angle C will be equal to 90 degrees. So this implies both angle A and angle C will be acute angles. So if any one of these trigonometric ratios of an acute angle same angle A is given then we can easily find out the remaining trigonometric ratios. Let's take one example for this suppose we are given cos A equal to 7 upon 25 and we need to find the remaining trigonometric ratios for the acute angle A. So for angle A BC is the perpendicular AB is the base and AC is the hypotenuse. Now cos A is B by H so 7 upon 25 is equal to B by H so this implies base that is B is equal to 7K and H that is hypotenuse is equal to 25K where K is any positive number. So here we have B equal to 7K and H equal to 25K and we need to find the value of perpendicular that is P. As it is a right angle triangle so to find the value of P that is perpendicular we can use the Pythagoras theorem. So by Pythagoras theorem we have P equal to square root of H square minus base that is B square and this will be equal to 25K square minus 7K square which will be equal to 24K. So we have P equal to 24K now let's find out the trigonometric ratios sin A is equal to P by H. Now the perpendicular is 24K so this will be 24K upon H that is 25K which will be equal to 24 by 25. As here K will get cancelled from the numerator and denominator. Now tan A is equal to P upon B that is perpendicular upon base so this will be equal to 24K upon 7K so this is equal to 24 upon 7. Now cosecant A is equal to 1 upon sin A so this will be equal to 25 upon 24 similarly secant A is equal to 1 upon cos A so this will be equal to 25 upon 7. And lastly cot A is equal to 1 upon tan A so this will be 7 upon 24. So here we were given one trigonometric ratio that is of cos A and we have found all the other trigonometric ratios. I hope this must be clear to you so with this we finish this session goodbye take care and have a nice day.