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A 16 m ladder is placed against the side of a house and reaches a window that is 12 m above the ground.1)How far from the base of the house is the ladder?(Round your answer to one decimal place and put only numbers in the blank with no units.)2) What is the angle of elevation of the ladder?(Round your answer to one decimal place and put only numbers in the blank with no units.)

Question

A 16 m ladder is placed against the side of a house and reaches a window that is 12 m above the ground.1)How far from the base of the house is the ladder?(Round your answer to one decimal place and put only numbers in the blank with no units.)2) What is the angle of elevation of the ladder?(Round your answer to one decimal place and put only numbers in the blank with no units.)

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Solution

  1. To find the distance from the base of the house to the ladder, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, the ladder is the hypotenuse, the height of the window is one side, and the distance from the base of the house to the ladder is the other side.

So, we have:

16^2 = 12^2 + x^2

256 = 144 + x^2

x^2 = 256 - 144

x^2 = 112

x = sqrt(112)

x = 10.6 (rounded to one decimal place)

So, the ladder is 10.6 m from the base of the house.

  1. To find the angle of elevation of the ladder, we can use the tangent function, which in a right triangle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In this case, the angle of elevation is the angle at the base of the ladder, the height of the window is the side opposite the angle, and the distance from the base of the house to the ladder is the side adjacent to the angle.

So, we have:

tan(theta) = 12 / 10.6

theta = arctan(12 / 10.6)

theta = 48.4 degrees (rounded to one decimal place)

So, the angle of elevation of the ladder is 48.4 degrees.

This problem has been solved

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