A ladder kept inclined at an angle of 40° with a distance of 8 feet from the building touches the building at some height. If the same ladder is kept at a distance of 6 feet with an inclination of 50° then find the height at which the ladder touches the building.
Answers
Answer:
Case 1:
Let the height at which the inclined ladder at an angle of 40° with the ground, touches the building be “h” feet. (As shown in the figure below).
Distance between the foot of the ladder and the building = 8 feet
According to the right-angled trigonometric ratio, we have
tan θ = (height of the building) / base
⇒ tan 40° = h / 8
⇒ h = 8 * 0.839 = 6.712 feet ….. (i)
Case 2:
Let the height at which the inclined ladder at an angle of 50° with the ground, touches the building be “(h + x)” feet. (As shown in the figure below).
Now, the distance between the foot of the ladder and the building = 6 feet
According to the right-angled trigonometric ratio, we have
tan θ = (height of the building) / base
⇒ tan 50° = (h+x) / 6
⇒ h+x = 6 * 1.191
⇒ h + x = 7.15
⇒ x = 7.15 - 6.712 ...... [from (i)]
⇒ x = 0.438 feet
∴ h + x = 6.712 + 0.438 = 7.15 feet
Thus, the height at which the ladder touches the building when ladder is inclined at an angle of 50° with the ground, is 7.15 feet.
