Asked by Alex
To measure the height of a building,two sightings are taken a distance of 50 feet apart. Of the first angle of elevation is 40 degrees and the second is 32 degrees, what is the height of the building?
Answers
Answered by
Reiny
Of course you made a diagram, right?
let's label A and B as the points where the two readings were taken, A being the 32 degree.
Let C be the base of the building and D the top of the building.
in triangle ABD , angle A = 32, angle ABD = 140 and angle ADB = 8
by sine law:
BD/sin32 = 50/sin8
BD = 50sin32/sin8 = 190.38
Now go back to the right-angled triangle BCD and use
sin 40 = DC/190.38
I am sure you can take it from here
let's label A and B as the points where the two readings were taken, A being the 32 degree.
Let C be the base of the building and D the top of the building.
in triangle ABD , angle A = 32, angle ABD = 140 and angle ADB = 8
by sine law:
BD/sin32 = 50/sin8
BD = 50sin32/sin8 = 190.38
Now go back to the right-angled triangle BCD and use
sin 40 = DC/190.38
I am sure you can take it from here
Answered by
albert
sin9 degere
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