Question

16. Mario is standing at ground level exactly at the corner where two exterior walls of his apartment building meet. From Mario’s position, his apartment window on the north side of the building appears 44.5 m away at an angle of elevation of Mario notices that his friend Thomas’s window on the west side of the building appears 71.0 m away at an angle of elevation of a) If a rope were pulled taut from one window to the other, around the outside of the building, how long, to the nearest tenth of a metre, would the rope need to be? Explain your reasoning. b) What is the straight-line distance through the building between the two windows? Round your answer to the nearest tenth of a metre.

Answer 1

a) To find the length of the rope around the outside of the building, we can use the law of cosines. Let be the length of the rope. According to the law of cosines:

c² = a² + b² - 2ab cos (C)

Here, a and b are distances from Mario to the north and west windows, respectively and C is the angle between these two directions.

Let us substitute the values:

c² = 44.5² + 71.0² - 2. 445 . 71.0 cos(90°)

Since 90° = 0, the equation is simplified to:

c² = 44.5² + 71.0²

c =
`\sqrt{44.5^(2) + 71.0^(2) }`

c = 1980.25 + 5041

c = 7021.25

c ≈ 83.79

b.