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Thread: How to find angle x marked in red? (bearings from Point X to pts R, S which are...)

  1. #1
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    How to find angle x marked in red? (bearings from Point X to pts R, S which are...)

    Hi, I don't have much of an attempt here, but here goes nothing:
    I'm trying to find the angle x marked in red, just trying to figure out methods using rules such as angles in a triangle, alternate etc.

    I have two bearings marked in orange from Point X to points R and S which are 270 and 210 degrees respectively.

    My thoughts are, angle RXS is actually 60 degrees, because of the bearing of S from X which is 210 degrees, and as you can see the 90 degree angle completes the full 360, so that is how I deduced this.

    Angle TSX is 25 degrees because it's an alternate angle of the angle of depression of S from T which is also 25 degrees.

    I'm racking my brain trying to see how I can single out x, I know it must be 25 minus something... but what?

    All angles are in degrees by the way.
    The question really asks to find the length of RS but I can do that easily once I have angle x in red.
    angles.jpg
    Last edited by richiesmasher; 02-08-2018 at 10:22 PM.

  2. #2
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    Quote Originally Posted by richiesmasher View Post
    Richie, that's gotta be your most confusing diagram so far

    Question:
    angleRTS = 20 degrees (shown in brown)
    That same angle is shown in blue as 35 - 25, thus 10 degrees
    Am I seeing things?!

    Curious: why show TX = 49m ?
    I'm just an imagination of your figment !

  3. #3
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    Quote Originally Posted by Denis View Post
    Richie, that's gotta be your most confusing diagram so far

    Question:
    angleRTS = 20 degrees (shown in brown)
    That same angle is shown in blue as 35 - 25, thus 10 degrees
    Am I seeing things?!

    Curious: why show TX = 49m ?
    No I actually wrote angle RTS is 10 degrees, perhaps it is difficult to see, well I showed it in case it had some relevance, I know exactly how to figure out the question, i just don't know how to find the angle i need to then utilize the sine rule.

  4. #4
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    OK; looked like 20...

    Had a more careful look using graph paper:

    R is on line SX, thus TXR and TXS are both right triangles.
    In other words, your x angle = 0 degrees!
    TXR = 90-55-35 and TXS = 90-65-25

    With TX = 49, Law of Sines makes it easy:
    calculate XR, then XS; SR = XS - XR

    That "210" simply shouldn't be there!
    I'm just an imagination of your figment !

  5. #5
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    Quote Originally Posted by Denis View Post
    OK; looked like 20...

    Had a more careful look using graph paper:

    R is on line SX, thus TXR and TXS are both right triangles.
    In other words, your x angle = 0 degrees!
    TXR = 90-55-35 and TXS = 90-65-25

    With TX = 49, Law of Sines makes it easy:
    calculate XR, then XS; SR = XS - XR

    That "210" simply shouldn't be there!
    Wow, I surprised myself, thanks dennis your clue made me see it clearly, it's two right angled triangles, from there I calculated XS and XR using tan of angles 25 and 35, then simply used the cosine rule using the angle I got from the bearing 210 and my answer 92m is correct ;D

  6. #6
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    I don't get that; what is equal to 92?
    I get XS = ~105 and XR = ~70, so RS = ~35

    But if you say 92 is the given answer, then I'm
    certainly missing something.
    It simply looks to me like the "210" appears in error.

    I sure hope someone else has a go at this problem.
    I'm just an imagination of your figment !

  7. #7
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    Quote Originally Posted by Denis View Post
    I don't get that; what is equal to 92?
    I get XS = ~105 and XR = ~70, so RS = ~35

    But if you say 92 is the given answer, then I'm
    certainly missing something.
    It simply looks to me like the "210" appears in error.

    I sure hope someone else has a go at this problem.
    20180209_094605.jpg
    Here is a better diagram, As you can see the triangle formed by the horizontal plane corresponds with the bearing 210. My reasoning was I need 360 degrees to complete the bearing, I already have 210 and I already have a 90, that gives me 300 degrees, so the only angle left to fill that space would be RXS which is 60 degrees, from there since i had the lengths of SX and RX, which were the same as yours, I applied the cosine rule because I had two sides and the angle inbetween.

    They now ask for the bearing of S from R, final part of the question, I'm currently racking my brain on this too, however we have more information now, I have 3 sides of the triangle RSX and one angle, I can possibly work out the other angles using sine rule and perhaps that can give me a clue.

  8. #8
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    Look at the diagram: you're saying that angleRXS = 60 degrees?
    Impossible.
    However, the word "bearing" gives me a headache.
    I quit here!
    Hope a better helper like DrP steps in.
    I'm just an imagination of your figment !

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