A helicopter is traveling at 86.0 km/h at an angle of 35° to the ground.

1. What is the value of [tex]\(A_x\)[/tex]? Round your answer to the nearest tenth.
[tex]\[ \square \text{ km/h} \][/tex]

2. What is the value of [tex]\(A_y\)[/tex]? Round your answer to the nearest tenth.
[tex]\[ \square \text{ km/h} \][/tex]



Answer :

Let's break down the problem step by step to find the horizontal and vertical components of the helicopter's speed, given its overall speed and angle of travel.

1. Understand the given values:
- Speed of the helicopter, [tex]\( V = 86.0 \)[/tex] km/h
- Angle of travel with respect to the ground, [tex]\( \theta = 35^\circ \)[/tex]

2. Identify what we need to find:
- The horizontal component of the speed, [tex]\( A_x \)[/tex]
- The vertical component of the speed, [tex]\( A_y \)[/tex]

3. Use trigonometric relationships:

The horizontal component [tex]\( A_x \)[/tex] can be found using the cosine function:
[tex]\[ A_x = V \cdot \cos(\theta) \][/tex]

The vertical component [tex]\( A_y \)[/tex] can be found using the sine function:
[tex]\[ A_y = V \cdot \sin(\theta) \][/tex]

4. Convert the angle from degrees to radians, as trigonometric functions generally use radians:
[tex]\[ \theta_{radians} = \theta_{degrees} \times \frac{\pi}{180} \][/tex]

5. Calculate the components:
- For [tex]\( A_x \)[/tex]:
[tex]\[ A_x = 86.0 \times \cos(35^\circ) = 70.4 \text{ km/h} \][/tex]

- For [tex]\( A_y \)[/tex]:
[tex]\[ A_y = 86.0 \times \sin(35^\circ) = 49.3 \text{ km/h} \][/tex]

6. Finally, round the answers to the nearest tenth.

Thus, the values of the components are:

- [tex]\( A_x = 70.4 \)[/tex] km/h
- [tex]\( A_y = 49.3 \)[/tex] km/h

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