Question

from a bridge 25m high the angel of depression of a boat is 45degree find the horizontal distance of the boat from the bridge? ​

Answer 1

Answer:

AB=BC=25m

AB=BC=25m∠B=90o

AB=BC=25m∠B=90o∠BAC=θ

AB=BC=25m∠B=90o∠BAC=θNow, in △ABC

AB=BC=25m∠B=90o∠BAC=θNow, in △ABC∠BAC=∠DCA  (alternate angle)

AB=BC=25m∠B=90o∠BAC=θNow, in △ABC∠BAC=∠DCA  (alternate angle)⇒tanθ=BCAB

AB=BC=25m∠B=90o∠BAC=θNow, in △ABC∠BAC=∠DCA  (alternate angle)⇒tanθ=BCAB⇒tanθ=1

AB=BC=25m∠B=90o∠BAC=θNow, in △ABC∠BAC=∠DCA  (alternate angle)⇒tanθ=BCAB⇒tanθ=1θ=45o  (∵tan450=1)

AB=BC=25m∠B=90o∠BAC=θNow, in △ABC∠BAC=∠DCA  (alternate angle)⇒tanθ=BCAB⇒tanθ=1θ=45o  (∵tan450=1)