From the top of a tower 40 m tall, the angle of depression of the nearer of the two points A and B on the ground on diametrically opposite sides of the tower is 45°. Find the angle of depression of the other point to the nearest degree if the distances of the two points from the base of the tower are in the ratio of 1:2
Answers
Answer:
Angle of elevation of the tower top from a point 40 m away from the foot of tower = 30° ==> h/40 = Tan 30° = 1/√3. On moving a distance of 20 metres towards the foot the tower to a point B, the angle of elevation increases to 60 0.
Answer:
The angle of depression of point B to the nearest degree is 60°.
Question : From the top of a tower 40 m tall, the angle of depression of the nearer of the two points A and B on the ground on diametrically opposite sides of the tower is 45°. Find the angle of depression of the other point to the nearest degree if the distances of the two points from the base of the tower are in the ratio of 1:2.
Step-by-step explanation:
From the above question,
They have given :
x = distance of point A from the base of the tower
2x = distance of point B from the base of the tower
Using the trigonometric ratios,
tan 45° = x/40
x = 40tan45°
= 28.3 m
2x = 56.6 m
tan θ = 56.6/40
θ = tan-1 (56.6/40) = 60°
Let x = distance of point A from the base of the tower
2x = distance of point B from the base of the tower
Draw a right triangle with the tower as the hypotenuse.
Since the angle of depression of point A is 45°,
tan 45° = x/40
x = 40 tan 45°
x = 40 × 1
x = 40 m
Therefore, the distance of point A from the base of the tower is 40 m.
Now, the distance of point B from the base of the tower is 2x,
2x = 2 × 40
2x = 80 m
Again, draw a right triangle with the tower as the hypotenuse.
Let θ be the angle of depression of point B to the nearest degree.
Using the trigonometric ratios,
tan θ = 80/40
θ = tan-1(80/40)
= 60°
Therefore, the angle of depression of point B to the nearest degree is 60°
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