Question
From the top of a tower the angle of depression of an object on the horizontal ground is found to be 60degree. On descending 20 m vertically downwards from the top of the tower, the angle of depression of the object is found to be 30degree Find the height of the tower.
Answers
Reiny
make your sketch
label the top of the tower A, bottom of tower B,
object on ground C and the point 20 m from the top as D
in triangle ACD
angle ACD = 60° - 35° = 25°
angle A = 180° - 90° - 60° = 30°
and AD = 20
by the sine law
CD/sin30 = 20/sin25
CD = 20sin30/sin25
in the right-angled triangle DCB
sin35 = DB/CD
DB = CDsin35
= 20sin30 sin35/sin25 , notice I have not yet touched my calculator , now I will
= 13.57
so the height of the tower is 20+13.57 = 33.57 m
label the top of the tower A, bottom of tower B,
object on ground C and the point 20 m from the top as D
in triangle ACD
angle ACD = 60° - 35° = 25°
angle A = 180° - 90° - 60° = 30°
and AD = 20
by the sine law
CD/sin30 = 20/sin25
CD = 20sin30/sin25
in the right-angled triangle DCB
sin35 = DB/CD
DB = CDsin35
= 20sin30 sin35/sin25 , notice I have not yet touched my calculator , now I will
= 13.57
so the height of the tower is 20+13.57 = 33.57 m
Steve
good work, but - where did the 35° come from?
Reiny
Rats!!
Just have to get better glasses, I read the 30° as 35°
Just how do you spot those little typos? Amazing.
Just have to get better glasses, I read the 30° as 35°
Just how do you spot those little typos? Amazing.