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