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Ludmilka [50]
2 years ago
10

Consider a triangle ABC

Mathematics
1 answer:
Sav [38]2 years ago
8 0

The given triangle has three angles with measurements: ∠A = 69°, ∠B = 52°, and ∠C = 59° respectively. Using the law of cosines, these angles are calculated from the given lengths of the triangle.

<h3>What is the law of cosines?</h3>

The law of cosines gives the relationship between the lengths of sides and the angles of the triangle ABC.

According to the law of cosines:

Cos A = (b² + c² - a²)/2bc

Cos B = (a² + c² - b²)/2ac

Cos C = (a² + b² - c²)/2ab

<h3>Calculation:</h3>

For the given triangle ABC,

a = 75, b = 63, and c = 69

So, using the law of cosines,

Cos A = (b² + c² - a²)/2bc

⇒ Cos A = (63² + 69² - 75²)/2×63×69

⇒ Cos A = 5/14

⇒ A = Cos⁻¹(5/14) = 69.07

∴ ∠A = 69°

Similarly,

Cos B = (a² + c² - b²)/2ac

⇒ Cos B = (75² + 69² - 63²)/2×75×69

⇒ Cos B = 31/50

⇒ B = Cos⁻¹(31/50) = 51.6 ≅ 52

∴ ∠B = 52°

Cos C = (a² + b² - c²)/2ab

⇒ Cos C = (75² + 63² - 69²)/2×75×63

⇒ Cos C = 179/350

⇒ C = Cos⁻¹(179/350) = 59.2

∴ C = 59°

Thus, the angles of the triangle ABC are 69°, 52°, and 59° respectively.

Learn more about the law of cosines here:

brainly.com/question/8288607

#SPJ1

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8 0
3 years ago
Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} A={1, 3, 4, 5, 7,
blagie [28]

Answer:

a) A' = [0,2,6,8]

b) (AUB)' = [0,2,6]

c) (AUB')' = [9]

d) A∩B′= [3,5,9]

Step-by-step explanation:

Assuming this problem: "Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} , and B={1, 4, 7, 8} . List the elemetns of the following sets in the increasing order: a) A′=  b) (A∪B)′={ , , }} c) (A∪B′)′={ }} d) A∩B′={ , , }}"

Part a

For this case we just need to find the elements in the universal set that are not in A. And we see that:

A' = [0,2,6,8]

And that represent the complement for A

Part b

For this case we need to find first the Union AUB who are the elements on A or B without repetition and we got:

AUB = [1,3,4,5,7,8,9]

And now the complement for (AUB)' are the elements that are not in AUB but are on the universal set and we got:

(AUB)' = [0,2,6]

Part c

For this case we need to find B' who are the elements on the universal set that are not in B

B' = [0,2,3,5,6,9]

Then we can find the union between AUB' and we got:

AUB' = [0,1,2,3,4,5,6,7,9]

And then the complment is just:

(AUB')' = [9]

Part d

For this case we just need to see the elements in common between A and B' and we got:

A∩B′= [3,5,9]

6 0
4 years ago
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