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Nuetrik [128]
3 years ago
12

Some help figuring out the answer?? Also explain a little how you got there

Mathematics
1 answer:
ipn [44]3 years ago
6 0

9514 1404 393

Answer:

  x = 10·cos(θ) -4·cot(θ)

Step-by-step explanation:

Apparently, we are to assume that the horizontal lines are parallel to each other.

The relevant trig relations are ...

  Sin = Opposite/Hypotenuse

  Cos = Adjacent/Hypotenuse

If the junction point in the middle of AB is labeled X, then we have ...

  sin(θ) = 4/BX   ⇒   BX = 4/sin(θ)

  cos(θ) = x/XA   ⇒   XA = x/cos(θ)

Then ...

  BX +XA = AB = 10

Substituting for BX and XA using the above relations, we get

  4/sin(θ) +x/cos(θ) = 10

Solving for x gives ...

  x = (10 -4/sin(θ))·cos(θ)

  x = 10·cos(θ) -4·cot(θ) . . . . . simplify

_____

We used the identity ...

   cot(θ) = cos(θ)/sin(θ)

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