 Hello and welcome to the session I am Shashi and I am going to help you with the following question. Question says the angles of elevation of the top of a tower from two points at the distance of four meters and nine meters from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is six meters. Let us now start the solution first of all we will draw a simple diagram to represent that problem. Now here AB is the tower now two points C and D are nine meters and four meters away from point B that is the base of the tower and angle of elevation of point A from point D is theta and angle of elevation of point A from point C is 90 minus theta we are given that these two angles are complementary angles we know complementary angles are those angles who have the sum 90 degrees so if one of the angles is theta then other angle is equal to 90 minus theta so we can write let AB is the tower and point C and D are at a distance of nine meters and four meters from point B that is the base of the tower now we are given that point C and point D are in the straight line and angle of elevation of A from these two points are complementary so we can write let angle AB is equal to theta that is angle of elevation of point A from point T this is equal to theta this implies angle ACB is equal to 90 minus theta since these two angles are complementary so angle ACB is equal to 90 minus theta now we have to prove that height of the tower is six meters that is we have to show that AB is equal to six meters now first of all let us consider right triangle ADB in right triangle ADB tan theta is equal to AB upon BD you know tan theta is equal to perpendicular upon base here AB is the perpendicular and BD is the base now we know BD is equal to four meters we are given that distance of point T from base of the tower that is point B is four meters so substituting BD is equal to four meters in this expression we get tan theta is equal to AB upon four let us name this expression as one now we will consider triangle ACB in right triangle ACB tan 90 minus theta is equal to AB upon BC we know tan theta is equal to perpendicular upon base in this triangle this is theta and this is perpendicular this is base so we get tan 90 minus theta is equal to AB upon BC now we know distance of point C from base of the tower that is point B is equal to nine meters or we can simply say BC is equal to nine meters now substituting value of BC in this expression we get tan 90 minus theta is equal to AB upon nine now we know tan 90 minus theta is equal to cot theta so this implies cot theta is equal to AB upon nine now let us name this expression as two now multiplying expressions one and two we get tan theta multiplied by cot theta is equal to AB upon four multiplied by AB upon nine now we know tan theta multiplied by one upon tan theta we know cot theta is reciprocal of tan theta is equal to AB square upon 36 now tan theta and tan theta will cancel each other and we get one is equal to AB square upon 36 now multiplying both the sides by 36 we get 36 is equal to AB square now taking square root on both the sides we get 6 is equal to AB we know square root of 36 is equal to plus minus 6 but height of the tower cannot be negative so we will neglect the negative value that is minus 6 and we get 6 is equal to AB or we can simply write it as AB is equal to 6 meters now we know AB is the height of the tower so height of the tower is equal to 6 meters hence proved this completes the session hope you understood the solution take care and have a nice day