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Nimfa-mama [501]
3 years ago
14

Can you help me with this plz

Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

A Q1

cos 59° = x/16

x = 16 cos 59°

x = 8.24

Q2

BC is given 23 mi

Maybe AB is needed

AB = √34² + 23² = 41 (rounded)

Q3

BC² = AB² - AC²

BC = √(37² - 12²) = 35

Q4

Let the angle is x

cos x = 19/20

x = arccos (19/20)

x = 18.2° (rounded)

Q5

See attached

Added point D and segments AD and DC to help with calculation

BC² = BD² + DC² = (AB + AD)² + DC²

Find the length of added red segments

AD = AC cos 65° = 14 cos 65° = 5.9

DC = AC sin 65° = 14 sin 65° = 12.7

Now we can find the value of BC

BC² = (19 + 5.9)² + 12.7²

BC = √781.3

BC = 28.0 yd

All calculations are rounded

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Answer:

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Step-by-step explanation:

It's both!

This is a might tricky. First you have to find the altitude. You have to determine if this is a real triangle.

Sin(22) = opposite / hypotenuse. The hypotenuse = 111. The angle is 22

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What have you learned?

Since 42 is larger than 41.58 you have 2 solutions to the triangle. One of the angles is acute, and the other one is obtuse. They are supplementary angles.

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