Part A:
Given that the cable runs from the top of a building to a point on the ground 60.3 feet from the base of the building and that the cable makes and angle of 33.7˚ with the ground.
Let the height of the building be h, then
[tex]\tan33.7^o= \frac{h}{60.3} \\ \\ \Rightarrow h=60.3\tan33.7^o \\ \\ \approx40\, feet.[/tex]
Part B:
Given that the cable runs from the top of a building to a point on the ground 60.3 feet from the base of the building and that the cable makes and angle of 33.7˚ with the ground.
Let the length of the building be l, then
[tex]\cos33.7^o= \frac{60.3}{l} \\ \\ \Rightarrow l=\frac{60.3}{\cos33.7^o} \\ \\ \approx72\, feet.[/tex]