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A man located at a point a on one bank of a river that is 1000 ft wide observed a woman jogging on the opposite bank. when the jogger was first spotted, the angle between the river bank and the man's line of sight was 25°. one minute later, the angle was 35°. how fast was the woman running if she maintained a constant speed? (round your answer to the nearest integer.)

2 Answers

6 votes

Final answer:

The woman is running at a speed of approximately 733 ft/min.

Step-by-step explanation:

To find the speed at which the woman is running, we can use the concept of trigonometry and the change in the angles over time. Let's assume the speed of the woman is v. Since the angle between the river bank and the line of sight decreases from 25° to 35° in 1 minute, we can set up the following equation: tan(35°) = (1000/v) - tan(25°).

Simplifying this equation, we get: v ≈ 733 ft/min. Rounding to the nearest integer, the speed of the woman running is approximately 733 ft/min.

User Sgriffin
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8.3k points
2 votes

Refer the given figure

Here A is the position of observer and B1 is the first position of lady and B2 is the position after one minute.

The distance traveled by lady = B1B2 = AC1 - AC2

We have AC1 = 1000
cot25^0 = 2144.51 m

AC2 = 1000
cot35^0 = 1428.15 m

The distance traveled by lady = B1B2 = AC1 - AC2 = 2144.51 - 1428.15 = 716.36 m

So velocity of lady = distance / time = 716.36/60 = 11.94 m/s ≈ 12 m/s

A man located at a point a on one bank of a river that is 1000 ft wide observed a-example-1
User James Thomas
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7.9k points