Math / Geometry

QuestionA triangular field has sides of lengths 24, 47, 59 km . Enter your answer as a number; answer should be accurate to 2 decimal places.
Find the largest angle in degrees: $\square$

Studdy Solution

STEP 1

What is this asking? We've got a huge triangular field, and we need to find the biggest angle inside it, knowing the lengths of all three sides! Watch out! Remember, the largest angle is always opposite the longest side.
Don't mix them up!

STEP 2

1. Apply Law of Cosines
2. Calculate the angle

STEP 3

The longest side is $59$ km.
This is super important because the largest angle will be opposite to this side.
Let's call this side $c$ , so $c = 59$ .

STEP 4

Let's call the other two sides $a$ and $b$ .
So, $a = 24$ and $b = 47$ .

STEP 5

The Law of Cosines states: $c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot \cos(C)$ , where $C$ is the angle opposite side $c$ .
This is exactly what we need!

STEP 6

Let's plug in what we know: $59^2 = 24^2 + 47^2 - 2 \cdot 24 \cdot 47 \cdot \cos(C)$ $3481 = 576 + 2209 - 2256 \cdot \cos(C)$

STEP 7

Let's simplify and isolate the cosine term: $3481 = 2785 - 2256 \cdot \cos(C)$ $3481 - 2785 = -2256 \cdot \cos(C)$ $696 = -2256 \cdot \cos(C)$

STEP 8

Divide both sides by $-2256$ to get $\cos(C)$ by itself: $\frac{696}{-2256} = \cos(C)$ $\cos(C) \approx -0.3085$

STEP 9

Now, we'll use the inverse cosine (also called arccos) to find the angle $C$ : $C = \arccos(-0.3085)$ $C \approx 107.94^\circ$

STEP 10

The largest angle in the triangular field is approximately 107.94 degrees.

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QuestionA triangular field has sides of lengths 24, 47, 59 km . Enter your answer as a number; answer should be accurate to 2 decimal places.
Find the largest angle in degrees: $\square$

Studdy Solution

STEP 1

What is this asking? We've got a huge triangular field, and we need to find the biggest angle inside it, knowing the lengths of all three sides! Watch out! Remember, the largest angle is always opposite the longest side.
Don't mix them up!

STEP 2

1. Apply Law of Cosines
2. Calculate the angle

STEP 3

The longest side is $59$ km.
This is super important because the largest angle will be opposite to this side.
Let's call this side $c$ , so $c = 59$ .

STEP 4

Let's call the other two sides $a$ and $b$ .
So, $a = 24$ and $b = 47$ .

STEP 5

The Law of Cosines states: $c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot \cos(C)$ , where $C$ is the angle opposite side $c$ .
This is exactly what we need!

STEP 6

Let's plug in what we know: $59^2 = 24^2 + 47^2 - 2 \cdot 24 \cdot 47 \cdot \cos(C)$ $3481 = 576 + 2209 - 2256 \cdot \cos(C)$

STEP 7

Let's simplify and isolate the cosine term: $3481 = 2785 - 2256 \cdot \cos(C)$ $3481 - 2785 = -2256 \cdot \cos(C)$ $696 = -2256 \cdot \cos(C)$

STEP 8

Divide both sides by $-2256$ to get $\cos(C)$ by itself: $\frac{696}{-2256} = \cos(C)$ $\cos(C) \approx -0.3085$

STEP 9

Now, we'll use the inverse cosine (also called arccos) to find the angle $C$ : $C = \arccos(-0.3085)$ $C \approx 107.94^\circ$

STEP 10

The largest angle in the triangular field is approximately 107.94 degrees.

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QuestionA triangular field has sides of lengths 24, 47, 59 km . Enter your answer as a number; answer should be accurate to 2 decimal places.Find the largest angle in degrees: □\square□

Studdy Solution

STEP 1

What is this asking? We've got a huge triangular field, and we need to find the biggest angle inside it, knowing the lengths of all three sides! Watch out! Remember, the largest angle is always opposite the longest side.Don't mix them up!

STEP 2

1. Apply Law of Cosines2. Calculate the angle

STEP 3

The longest side is 595959 km.This is super important because the largest angle will be *opposite* to this side.Let's call this side ccc, so c=59c = 59c=59.

STEP 4

Let's call the other two sides aaa and bbb.So, a=24a = 24a=24 and b=47b = 47b=47.

STEP 5

The Law of Cosines states: c2=a2+b2−2⋅a⋅b⋅cos⁡(C)c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot \cos(C)c2=a2+b2−2⋅a⋅b⋅cos(C), where CCC is the angle opposite side ccc.This is *exactly* what we need!

STEP 6

Let's plug in what we know: 592=242+472−2⋅24⋅47⋅cos⁡(C)59^2 = 24^2 + 47^2 - 2 \cdot 24 \cdot 47 \cdot \cos(C)592=242+472−2⋅24⋅47⋅cos(C) 3481=576+2209−2256⋅cos⁡(C)3481 = 576 + 2209 - 2256 \cdot \cos(C)3481=576+2209−2256⋅cos(C)

STEP 7

Let's simplify and isolate the cosine term: 3481=2785−2256⋅cos⁡(C)3481 = 2785 - 2256 \cdot \cos(C)3481=2785−2256⋅cos(C) 3481−2785=−2256⋅cos⁡(C)3481 - 2785 = -2256 \cdot \cos(C)3481−2785=−2256⋅cos(C)696=−2256⋅cos⁡(C)696 = -2256 \cdot \cos(C)696=−2256⋅cos(C)

STEP 8

Divide both sides by −2256-2256−2256 to get cos⁡(C)\cos(C)cos(C) by itself: 696−2256=cos⁡(C)\frac{696}{-2256} = \cos(C)−2256696​=cos(C) cos⁡(C)≈−0.3085\cos(C) \approx -0.3085cos(C)≈−0.3085

STEP 9

Now, we'll use the inverse cosine (also called arccos) to find the angle CCC: C=arccos⁡(−0.3085)C = \arccos(-0.3085)C=arccos(−0.3085) C≈107.94∘C \approx 107.94^\circC≈107.94∘

STEP 10

The largest angle in the triangular field is approximately **107.94** degrees.

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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.

QuestionA triangular field has sides of lengths 24, 47, 59 km . Enter your answer as a number; answer should be accurate to 2 decimal places.
Find the largest angle in degrees: $\square$

What is this asking? We've got a huge triangular field, and we need to find the biggest angle inside it, knowing the lengths of all three sides! Watch out! Remember, the largest angle is always opposite the longest side.
Don't mix them up!

1. Apply Law of Cosines
2. Calculate the angle

The longest side is $59$ km.
This is super important because the largest angle will be opposite to this side.
Let's call this side $c$ , so $c = 59$ .

Let's call the other two sides $a$ and $b$ .
So, $a = 24$ and $b = 47$ .

The Law of Cosines states: $c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot \cos(C)$ , where $C$ is the angle opposite side $c$ .
This is exactly what we need!

Let's plug in what we know: $59^2 = 24^2 + 47^2 - 2 \cdot 24 \cdot 47 \cdot \cos(C)$ $3481 = 576 + 2209 - 2256 \cdot \cos(C)$

Let's simplify and isolate the cosine term: $3481 = 2785 - 2256 \cdot \cos(C)$ $3481 - 2785 = -2256 \cdot \cos(C)$ $696 = -2256 \cdot \cos(C)$

Divide both sides by $-2256$ to get $\cos(C)$ by itself: $\frac{696}{-2256} = \cos(C)$ $\cos(C) \approx -0.3085$

Now, we'll use the inverse cosine (also called arccos) to find the angle $C$ : $C = \arccos(-0.3085)$ $C \approx 107.94^\circ$

The largest angle in the triangular field is approximately 107.94 degrees.