Math / Trigonometry

QuestionYou are on top of a 40 m high cliff. You see two boats in the water. Boat A is at an anlge depression of $15^{\circ}$ . Boat B is at an angle of depression of $27^{\circ}$ . Find the distance between th two boats.

Studdy Solution

STEP 1

What is this asking? We need to find the distance between two boats viewed from a cliff, knowing the cliff's height and the angles at which we see the boats. Watch out! Don't mix up angles of depression with angles of elevation!
Also, remember SOH CAH TOA.

STEP 2

1. Visualize and label
2. Find the distance to Boat A
3. Find the distance to Boat B
4. Calculate the distance between the boats

STEP 3

Let's draw a diagram!
We have a cliff that's $40$ m high.
We'll label that.
We have two boats, A and B.
The angle of depression to Boat A is $15^\circ$ , and the angle of depression to Boat B is $27^\circ$ .
Remember, the angle of depression is the angle downwards from the horizontal.
Let's label those angles!

STEP 4

We want to find the distance between the boats.
Let's call the distance from the base of the cliff to Boat A, $x$ , and the distance from the base of the cliff to Boat B, $y$ .
The distance between the boats will then be $x - y$ .

STEP 5

We have a right triangle with the height of the cliff ( $40$ m) and the distance $x$ .
We know the angle of depression is $15^\circ$ , which means the angle inside our triangle is also $15^\circ$ .
We can use tangent!
Remember SOH CAH TOA?
Tangent is opposite over adjacent, so $\tan(15^\circ) = \frac{40}{x}$ .

STEP 6

To solve for $x$ , we can multiply both sides by $x$ and then divide both sides by $\tan(15^\circ)$ .
This gives us $x = \frac{40}{\tan(15^\circ)}$ .

STEP 7

Calculating this gives us $x \approx \frac{40}{0.268} \approx 149.25$ .
So, $x \approx \textbf{149.25}$ m.

STEP 8

Now, let's do the same thing for Boat B!
We have another right triangle, this time with the height of the cliff ( $40$ m) and the distance $y$ .
The angle inside the triangle is $27^\circ$ .
So, $\tan(27^\circ) = \frac{40}{y}$ .

STEP 9

Just like before, we solve for $y$ : $y = \frac{40}{\tan(27^\circ)}$ .

STEP 10

Calculating this gives us $y \approx \frac{40}{0.510} \approx 78.43$ .
So, $y \approx \textbf{78.43}$ m.

STEP 11

The distance between the boats is simply $x - y$ .

STEP 12

We found that $x \approx 149.25$ m and $y \approx 78.43$ m.
Therefore, the distance between the boats is approximately $149.25 - 78.43 = \textbf{70.82}$ m.

STEP 13

The distance between the two boats is approximately $\textbf{70.82}$ meters.

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QuestionYou are on top of a 40 m high cliff. You see two boats in the water. Boat A is at an anlge depression of $15^{\circ}$ . Boat B is at an angle of depression of $27^{\circ}$ . Find the distance between th two boats.

Studdy Solution

STEP 1

What is this asking? We need to find the distance between two boats viewed from a cliff, knowing the cliff's height and the angles at which we see the boats. Watch out! Don't mix up angles of depression with angles of elevation!
Also, remember SOH CAH TOA.

STEP 2

1. Visualize and label
2. Find the distance to Boat A
3. Find the distance to Boat B
4. Calculate the distance between the boats

STEP 3

STEP 4

We want to find the distance between the boats.
Let's call the distance from the base of the cliff to Boat A, $x$ , and the distance from the base of the cliff to Boat B, $y$ .
The distance between the boats will then be $x - y$ .

STEP 5

STEP 6

To solve for $x$ , we can multiply both sides by $x$ and then divide both sides by $\tan(15^\circ)$ .
This gives us $x = \frac{40}{\tan(15^\circ)}$ .

STEP 7

Calculating this gives us $x \approx \frac{40}{0.268} \approx 149.25$ .
So, $x \approx \textbf{149.25}$ m.

STEP 8

Now, let's do the same thing for Boat B!
We have another right triangle, this time with the height of the cliff ( $40$ m) and the distance $y$ .
The angle inside the triangle is $27^\circ$ .
So, $\tan(27^\circ) = \frac{40}{y}$ .

STEP 9

Just like before, we solve for $y$ : $y = \frac{40}{\tan(27^\circ)}$ .

STEP 10

Calculating this gives us $y \approx \frac{40}{0.510} \approx 78.43$ .
So, $y \approx \textbf{78.43}$ m.

STEP 11

The distance between the boats is simply $x - y$ .

STEP 12

We found that $x \approx 149.25$ m and $y \approx 78.43$ m.
Therefore, the distance between the boats is approximately $149.25 - 78.43 = \textbf{70.82}$ m.

STEP 13

The distance between the two boats is approximately $\textbf{70.82}$ meters.

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QuestionYou are on top of a 40 m high cliff. You see two boats in the water. Boat A is at an anlge depression of 15∘15^{\circ}15∘. Boat B is at an angle of depression of 27∘27^{\circ}27∘. Find the distance between th two boats.

Studdy Solution

STEP 1

What is this asking? We need to find the distance between two boats viewed from a cliff, knowing the cliff's height and the angles at which we see the boats. Watch out! Don't mix up angles of depression with angles of elevation!Also, remember SOH CAH TOA.

STEP 2

1. Visualize and label2. Find the distance to Boat A3. Find the distance to Boat B4. Calculate the distance between the boats

STEP 3

STEP 4

We want to find the distance between the boats.Let's call the distance from the base of the cliff to Boat A, xxx, and the distance from the base of the cliff to Boat B, yyy.The distance between the boats will then be x−yx - yx−y.

STEP 5

STEP 6

To solve for xxx, we can multiply both sides by xxx and then divide both sides by tan⁡(15∘)\tan(15^\circ)tan(15∘).This gives us x=40tan⁡(15∘)x = \frac{40}{\tan(15^\circ)}x=tan(15∘)40​.

STEP 7

Calculating this gives us x≈400.268≈149.25x \approx \frac{40}{0.268} \approx 149.25x≈0.26840​≈149.25.So, x≈149.25x \approx \textbf{149.25}x≈149.25 m.

STEP 8

Now, let's do the same thing for Boat B!We have another right triangle, this time with the height of the cliff (404040 m) and the distance yyy.The angle inside the triangle is 27∘27^\circ27∘.So, tan⁡(27∘)=40y\tan(27^\circ) = \frac{40}{y}tan(27∘)=y40​.

STEP 9

Just like before, we solve for yyy: y=40tan⁡(27∘)y = \frac{40}{\tan(27^\circ)}y=tan(27∘)40​.

STEP 10

Calculating this gives us y≈400.510≈78.43y \approx \frac{40}{0.510} \approx 78.43y≈0.51040​≈78.43.So, y≈78.43y \approx \textbf{78.43}y≈78.43 m.

STEP 11

The distance between the boats is simply x−yx - yx−y.

STEP 12

We found that x≈149.25x \approx 149.25x≈149.25 m and y≈78.43y \approx 78.43y≈78.43 m.Therefore, the distance between the boats is approximately 149.25−78.43=70.82149.25 - 78.43 = \textbf{70.82}149.25−78.43=70.82 m.

STEP 13

The distance between the two boats is approximately 70.82\textbf{70.82}70.82 meters.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionYou are on top of a 40 m high cliff. You see two boats in the water. Boat A is at an anlge depression of $15^{\circ}$ . Boat B is at an angle of depression of $27^{\circ}$ . Find the distance between th two boats.

What is this asking? We need to find the distance between two boats viewed from a cliff, knowing the cliff's height and the angles at which we see the boats. Watch out! Don't mix up angles of depression with angles of elevation!
Also, remember SOH CAH TOA.

1. Visualize and label
2. Find the distance to Boat A
3. Find the distance to Boat B
4. Calculate the distance between the boats

We want to find the distance between the boats.
Let's call the distance from the base of the cliff to Boat A, $x$ , and the distance from the base of the cliff to Boat B, $y$ .
The distance between the boats will then be $x - y$ .

To solve for $x$ , we can multiply both sides by $x$ and then divide both sides by $\tan(15^\circ)$ .
This gives us $x = \frac{40}{\tan(15^\circ)}$ .

Calculating this gives us $x \approx \frac{40}{0.268} \approx 149.25$ .
So, $x \approx \textbf{149.25}$ m.

Now, let's do the same thing for Boat B!
We have another right triangle, this time with the height of the cliff ( $40$ m) and the distance $y$ .
The angle inside the triangle is $27^\circ$ .
So, $\tan(27^\circ) = \frac{40}{y}$ .

Just like before, we solve for $y$ : $y = \frac{40}{\tan(27^\circ)}$ .

Calculating this gives us $y \approx \frac{40}{0.510} \approx 78.43$ .
So, $y \approx \textbf{78.43}$ m.

The distance between the boats is simply $x - y$ .

We found that $x \approx 149.25$ m and $y \approx 78.43$ m.
Therefore, the distance between the boats is approximately $149.25 - 78.43 = \textbf{70.82}$ m.

The distance between the two boats is approximately $\textbf{70.82}$ meters.