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2 years ago

1. A boat drifted 15 m out into the water off Ocean Beach, where the shore drops off

Mathematics

1 answer:

Lunna [17]2 years ago

7 0

The correct responses found using trigonometric ratios are as follows;

Brooke's cannot bring the boat back without getting water over her boots.
The plan will not be approved
The plane will not disturb
The company can not use the chimney legally
The Long company will build the bridge
The fireman can reach her.
The lift does not take the skiers up to the bar.
Eric will be able to see his Dad's plane.
The 73 m. brace

<h3>How can trigonometric ratios be used to find the right responses?</h3>

1. The depth of the water = 15 × tan(4°) = 1.049 m

The height of Brooke's boot = 1 m

Therefore;

Brooke's cannot bring the boat back without getting water over her boots.

2. The angle of elevation of the plan = arccos(99/100) ≈ 8.110°

Given that the angle of elevation the plan, 8.110°, is more than 8° limit, we find;

The plan will not be approved.

3. Height reached by the plane = 9 × tan(21°) = 3.455

Therefore;

The plane will not disturb.

4. Height of the chimney at the factory = 50 × tan(51°) = 61.745

The minimum chimney height = 65 m.

Therefore;

The company can not use the chimney legally.

6. Length of the bridge = 72 × tan(55°) ≈ 102.83

Therefore;

The Long company will build the bridge because it is longer than 100 m. but not more than 103 m.

7. Height reached by the ladder = 26 × sin(43°) ≈ 17.7

Therefore;

The fireman can reach her.

8. Height reached by the ski lift = 1000 × sin(29°) ≈ 484.8

Therefore;

The lift does not take the skiers up to the bar.

9. Height reached by plane = 26 × tan(9°) ≈ 4.12

The plane flies under the cloud therefore;

Eric will be able to see his Dad's plane.

10. Length required for the metal brace = 21/(cos(73°)) ≈ 71.83

Therefore;

The brace that wastes less metal is the 73 m. brace, because the 71 m. brace is not long enough and two braces will be required.

Learn more about trigonometric ratios here:

brainly.com/question/24349828

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Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh

yuradex [85]

Answer:

The northern lighthouse is approximately $24.4\; \rm mi$ closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as $\rm N$ , the southern lighthouse as $\rm S$ , and the boat as $\rm B$ . These three points would form a triangle.

It is given that two of the angles of this triangle measure $40^{\circ}$ (northern lighthouse, $\angle {\rm N}$ ) and $21^{\circ}$ (southern lighthouse $\angle {\rm S}$ ), respectively. The three angles of any triangle add up to $180^{\circ}$ . Therefore, the third angle of this triangle would measure $180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}$ (boat $\angle {\rm B}$ .)

It is also given that the length between the two lighthouses (length of $\rm NS$ ) is $75\; \rm mi$ .

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, $\angle {\rm B}$ is opposite to side $\rm NS$ , whereas $\angle {\rm S}$ is opposite to side ${\rm NB}$ . Therefore:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}$ .

Substitute in the known measurements:

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$ .

Rearrange and solve for the length of $\rm NB$ :

$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

(Round to at least one more decimal places than the values in the choices.)

Likewise, with $\angle {\rm N}$ is opposite to side ${\rm SB}$ , the following would also hold:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$ .

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$ .

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

In other words, the distance between the northern lighthouse and the boat is approximately $30.73\; \rm mi$ , whereas the distance between the southern lighthouse and the boat is approximately $55.12\; \rm mi$ . Hence the conclusion.