 Hi and welcome to the session, I am Deepika here. Let's discuss the question given 15 part A is equal to 8, find sine A and secant A. Now we know that according to the Pythagoras in triangle ABC, right angle at B is equal to AB square plus BC square that is hypotenuse square is equal to sum of the squares of other two sides and cot A is equal to side adjacent to angle upon side opposite to angle A and sine A is equal to opposite to angle A and hypotenuse is equal to fuse adjacent to angle A. So this is the key idea behind our question. I will pop this key idea to solve our question. So let's start the solution given 15 cot A is equal to 8. Therefore cot A is equal to 8 upon 15. ABC is the right angle at B. Now we know that cot A is equal to side adjacent to angle A upon side opposite to angle A that is cot A is equal to AB upon BC and this is given to us 8 over 15. Therefore if AB is equal to 8x then BC is equal to 15x where x is any positive number that is this is 8x then it is equal to 15x. Now according to the Pythagoras theorem it is equal to AB square plus BC square. Now AC square is equal to AB square is 8x square plus BC square that is 15x square. So this is equal to 25x square. This implies AC square is equal to C is equal to under root of 289x square which is equal to 17x. Hence ACs are 17x. Therefore sine A is equal to upon AC and this is equal to 15x upon 17x which is equal to 15 over 17 is equal to B upon AB that is 17x upon 8x and this is equal to 17 over 8. Hence the answer for the above question is sine A is equal to 15 over 17 and secant A is equal to 17 over 8. I hope this solution is dear to you. Bye and take care.