Math / Geometry

QuestionTwo sides of a sloped ceiling meet at an angle of $115.6^{\circ}$ . If the distances along the sides to the opposite walls are 11.5 ft and 14.7 ft , what length of bearn is needed to join the walls?
A $\square$ ft beam is needed to join the walls. (Round to the nearest tenth as needed.)

Studdy Solution

STEP 1

What is this asking? We need to find the length of a beam needed to span between two walls that meet at an angle, given the distances along the sloped ceiling to each wall. Watch out! Don't forget to convert the angle to radians if your cosine function requires it, and remember we're dealing with a triangle that isn't a right triangle!

STEP 2

1. Visualize and Define
2. Apply Law of Cosines
3. Calculate and Round

STEP 3

Alright, let's imagine this!
We've got two walls and a sloped ceiling forming a triangle.
The lengths along the ceiling to the walls are like two sides of our triangle, and the beam we need is the third side, connecting the walls.

STEP 4

Let's define $c$ as the length of the beam we're looking for.
We'll call the given lengths along the ceiling $a = 11.5$ ft and $b = 14.7$ ft.
The angle between the ceiling sides, which we'll call $\gamma$ , is 115.6 degrees.

STEP 5

The Law of Cosines is perfect for this!
It relates the sides and angles of any triangle, not just right triangles.
It states: $c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot \cos(\gamma)$ This tells us that the square of the side opposite an angle is equal to the sum of the squares of the other two sides minus twice their product times the cosine of the angle.

STEP 6

We need $c$ , the length of the beam, so let's plug in our values: $c^2 = (11.5)^2 + (14.7)^2 - 2 \cdot (11.5) \cdot (14.7) \cdot \cos(115.6^\circ)$

STEP 7

Time to crunch the numbers!
First, let's square $a$ and $b$ : $(11.5)^2 = 132.25$ $(14.7)^2 = 216.09$ So, $c^2 = 132.25 + 216.09 - 2 \cdot (11.5) \cdot (14.7) \cdot \cos(115.6^\circ)$

STEP 8

Now, let's calculate that cosine value: $\cos(115.6^\circ) \approx -0.4226$ So, $c^2 = 132.25 + 216.09 - 2 \cdot (11.5) \cdot (14.7) \cdot (-0.4226)$

STEP 9

Let's multiply and add everything together: $c^2 = 132.25 + 216.09 + 141.20$ $c^2 \approx 489.54$

STEP 10

Finally, take the square root to find $c$ : $c = \sqrt{489.54} \approx 22.1$ Remember to round to the nearest tenth!

STEP 11

A 22.1 ft beam is needed to join the walls.

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QuestionTwo sides of a sloped ceiling meet at an angle of $115.6^{\circ}$ . If the distances along the sides to the opposite walls are 11.5 ft and 14.7 ft , what length of bearn is needed to join the walls?
A $\square$ ft beam is needed to join the walls. (Round to the nearest tenth as needed.)

Studdy Solution

STEP 1

STEP 2

1. Visualize and Define
2. Apply Law of Cosines
3. Calculate and Round

STEP 3

Alright, let's imagine this!
We've got two walls and a sloped ceiling forming a triangle.
The lengths along the ceiling to the walls are like two sides of our triangle, and the beam we need is the third side, connecting the walls.

STEP 4

Let's define $c$ as the length of the beam we're looking for.
We'll call the given lengths along the ceiling $a = 11.5$ ft and $b = 14.7$ ft.
The angle between the ceiling sides, which we'll call $\gamma$ , is 115.6 degrees.

STEP 5

STEP 6

We need $c$ , the length of the beam, so let's plug in our values: $c^2 = (11.5)^2 + (14.7)^2 - 2 \cdot (11.5) \cdot (14.7) \cdot \cos(115.6^\circ)$

STEP 7

Time to crunch the numbers!
First, let's square $a$ and $b$ : $(11.5)^2 = 132.25$ $(14.7)^2 = 216.09$ So, $c^2 = 132.25 + 216.09 - 2 \cdot (11.5) \cdot (14.7) \cdot \cos(115.6^\circ)$

STEP 8

Now, let's calculate that cosine value: $\cos(115.6^\circ) \approx -0.4226$ So, $c^2 = 132.25 + 216.09 - 2 \cdot (11.5) \cdot (14.7) \cdot (-0.4226)$

STEP 9

Let's multiply and add everything together: $c^2 = 132.25 + 216.09 + 141.20$ $c^2 \approx 489.54$

STEP 10

Finally, take the square root to find $c$ : $c = \sqrt{489.54} \approx 22.1$ Remember to round to the nearest tenth!

STEP 11

A 22.1 ft beam is needed to join the walls.

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Studdy Solution

STEP 1

STEP 2

1. Visualize and Define2. Apply Law of Cosines3. Calculate and Round

STEP 3

Alright, let's **imagine** this!We've got two walls and a sloped ceiling forming a triangle.The lengths along the ceiling to the walls are like two sides of our triangle, and the beam we need is the third side, connecting the walls.

STEP 4

Let's **define** ccc as the length of the beam we're looking for.We'll call the given lengths along the ceiling a=11.5a = 11.5a=11.5 ft and b=14.7b = 14.7b=14.7 ft.The angle between the ceiling sides, which we'll call γ\gammaγ, is **115.6** degrees.

STEP 5

STEP 6

We need ccc, the length of the beam, so let's **plug in** our values: c2=(11.5)2+(14.7)2−2⋅(11.5)⋅(14.7)⋅cos⁡(115.6∘)c^2 = (11.5)^2 + (14.7)^2 - 2 \cdot (11.5) \cdot (14.7) \cdot \cos(115.6^\circ)c2=(11.5)2+(14.7)2−2⋅(11.5)⋅(14.7)⋅cos(115.6∘)

STEP 7

STEP 8

STEP 9

Let's **multiply** and **add** everything together: c2=132.25+216.09+141.20c^2 = 132.25 + 216.09 + 141.20c2=132.25+216.09+141.20 c2≈489.54c^2 \approx 489.54c2≈489.54

STEP 10

Finally, take the **square root** to find ccc: c=489.54≈22.1c = \sqrt{489.54} \approx 22.1c=489.54​≈22.1 Remember to round to the nearest tenth!

STEP 11

A **22.1** ft beam is needed to join the walls.

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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.

1. Visualize and Define
2. Apply Law of Cosines
3. Calculate and Round

Alright, let's imagine this!
We've got two walls and a sloped ceiling forming a triangle.
The lengths along the ceiling to the walls are like two sides of our triangle, and the beam we need is the third side, connecting the walls.

Let's define $c$ as the length of the beam we're looking for.
We'll call the given lengths along the ceiling $a = 11.5$ ft and $b = 14.7$ ft.
The angle between the ceiling sides, which we'll call $\gamma$ , is 115.6 degrees.

We need $c$ , the length of the beam, so let's plug in our values: $c^2 = (11.5)^2 + (14.7)^2 - 2 \cdot (11.5) \cdot (14.7) \cdot \cos(115.6^\circ)$

Let's multiply and add everything together: $c^2 = 132.25 + 216.09 + 141.20$ $c^2 \approx 489.54$

Finally, take the square root to find $c$ : $c = \sqrt{489.54} \approx 22.1$ Remember to round to the nearest tenth!

A 22.1 ft beam is needed to join the walls.