Help with these 3 I did all the other questions of the assignment, these are confusing ASAP

Answers:

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Explanation:

The diagrams for problems 7 and 8 are shown below.

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Problem 7

Apply the tangent ratio to find that...

tan(angle) = opposite/adjacent

tan(42) = x/28

28*tan(42) = x

x = 28*tan(42)

x = 25.2113132403396

x = 25.2

Make sure your calculator is in degree mode. One way to check is to type in tan(45) and you should get 1 as a result.

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Problem 8, part 1

To find the angle y, we must use the sine ratio

sin(angle) = opposite/hypotenuse

sin(y) = 18/23

y = arcsin(18/23)

y = 51.5000495907521

y = 52 degrees approximately

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Problem 8, part 2

We can use the pythagorean theorem to find the missing side x

a^2+b^2 = c^2

18^2+x^2 = 23^2

324+x^2 = 529

x^2 = 529-324

x^2 = 205

x = sqrt(205)

x = 14.3178210632763

x = 14.3

Or alternatively, we can apply the tangent ratio on the angle y we found earlier to help find x

tan(angle) = opposite/adjacent

tan(y) = 18/x

tan(51.50004959) = 18/x

x*tan(51.50004959) = 18

x = 18/tan(51.50004959)

x = 14.3178210636621

x = 14.3

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Problem 9

The double tickmarks show that BE = EC = 10. That means BC = BE+EC = 10+10 = 20.

Since we have similar triangles, we can solve for x like so

DE/AC = BE/BC

x/44 = 10/20

20x = 44*10

20x = 440

x = 440/20

x = 22

So DE is 22 units long. Note how this is half as long as the side AC = 44.

you would use (d.) the law of cosines or a^2 = b^2 + c^2 – 2bc(cos a)

, b^2 = a^2 + c^2 – 2ac(cos b), or c^2 = a^2 + b^2 - 2ab(cos c). you wouldn't use b. the law of sines as it is mostly effective when 1side and 2angles are given. a and c aren't the answer for they are the improper versions of the law of cosines. hope this is .

step-by-step explanation: