Question
2. A person walking along a straight road observed that at two consecutive kilometre stones the angles of elevation of a hill front of him are 40o and 80o respectively. Find the height of the hill.
Answers
Make a sketch
I have A, B, and C on a horizontal line, and H as the top of the hill
Angle A = 40° and angle HBC - 80°, AB = 2 km
In triangle ABH, angle ABH = 100°
then angle AHB = 40°
which makes BH = 2 , (isosceles triangle)
(that was lucky, else I would need the sine law to find BH)
in triangle HBC (right-angled)
sin 80 = HC/2
HC = 2sin80 = 1.97 km
Wow, that is more than just a "hill"
I have A, B, and C on a horizontal line, and H as the top of the hill
Angle A = 40° and angle HBC - 80°, AB = 2 km
In triangle ABH, angle ABH = 100°
then angle AHB = 40°
which makes BH = 2 , (isosceles triangle)
(that was lucky, else I would need the sine law to find BH)
in triangle HBC (right-angled)
sin 80 = HC/2
HC = 2sin80 = 1.97 km
Wow, that is more than just a "hill"
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