Results 1 to 3 of 3

Math Help - trigo (3)

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    trigo (3)

    A surveyor at a point A observes that the peak M of a mountain lies in the direction \phi degree east of north and the angle of elevation of AM above the horizontal is \alpha degree . He also observes that a point B lies due north of A and the angle of elevation of AB above the horizontal is \theta . The surveyor moves to the point B and observes that M now lies in the direction \omega east of north . He measures that the distance AB and finds that it is equal to l . Show that the height of M above the horizontal which passes through A is

     <br />
\frac{l\tan\alpha\sin\omega\cos\theta}{\sin(\omega-\phi)}<br />

    Show also that the angle of elevation \beta of BM above the horizontal is given by

     <br />
\tan\beta =\csc \phi[\tan\alpha\sin\omega-\tan\theta\sin(\omega-\phi)]<br />


    I am unsure about the sketch . Can someone show it to me because i cant do anything without the sketch . Really thanks for your time !! Can someone correct my latex as well . Thanks .
    Last edited by Chris L T521; August 13th 2009 at 03:49 AM. Reason: fixed LaTeX.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    3D Trigonometry

    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    A surveyor at a point A observes that the peak M of a mountain lies in the direction \phi degree east of north and the angle of elevation of AM above the horizontal is \alpha degree . He also observes that a point B lies due north of A and the angle of elevation of AB above the horizontal is \theta . The surveyor moves to the point B and observes that M now lies in the direction \omega east of north . He measures that the distance AB and finds that it is equal to l . Show that the height of M above the horizontal which passes through A is

     <br />
\frac{l\tan\alpha\sin\omega\cos\theta}{\sin(\omega-\phi)}<br />

    Show also that the angle of elevation \beta of BM above the horizontal is given by

     <br />
\tan\beta =\csc \phi[\tan\alpha\sin\omega-\tan\theta\sin(\omega-\phi)]<br />


    I am unsure about the sketch . Can someone show it to me because i cant do anything without the sketch . Really thanks for your time !! Can someone correct my latex as well . Thanks .
    Have a look at the sketch I've attached.

    BD and MC are vertical. ADN is the horizontal north-line through A. AC is horizontal.

    Then \angle ADB = \angle ACM = 90^o,\,\angle BAD = \theta,\, \angle DAC = \phi,\, \angle NDC = \omega,\, \angle MAC = \alpha

    The distance AB = l.

    Suppose the height MC = h. Use the right-angled triangles to write down, in terms of l and h (and sines/cosines\tangents of various angles) the lengths of AD and AC. Then use the Sine Rule on the \triangle ACD, and you're there.

    For part 2, draw a horizontal line through B to meet MC at E. Therefore \angle MBE = \beta.

    Then work out the distances BE \,(= DC) and ME in terms of l. Then use \tan\beta = \frac{ME}{BE} to get the result.

    You can do it!

    Grandad
    Attached Thumbnails Attached Thumbnails trigo (3)-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Quote Originally Posted by Grandad View Post
    Hello thereddevilsHave a look at the sketch I've attached.

    BD and MC are vertical. ADN is the horizontal north-line through A. AC is horizontal.

    Then \angle ADB = \angle ACM = 90^o,\,\angle BAD = \theta,\, \angle DAC = \phi,\, \angle NDC = \omega,\, \angle MAC = \alpha

    The distance AB = l.

    Suppose the height MC = h. Use the right-angled triangles to write down, in terms of l and h (and sines/cosines\tangents of various angles) the lengths of AD and AC. Then use the Sine Rule on the \triangle ACD, and you're there.

    For part 2, draw a horizontal line through B to meet MC at E. Therefore \angle MBE = \beta.

    Then work out the distances BE \,(= DC) and ME in terms of l. Then use \tan\beta = \frac{ME}{BE} to get the result.

    You can do it!

    Grandad
    I thank you for ur great effort , Grandad .. Thanks!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigo Eqn
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: December 2nd 2009, 02:53 AM
  2. Trigo qn
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: October 8th 2008, 04:13 AM
  3. Trigo qn
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 7th 2008, 04:39 AM
  4. trigo sum/different~do not do~!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: September 11th 2008, 06:43 AM
  5. trigo > i cant do it!
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 9th 2008, 05:43 AM

Search Tags


/mathhelpforum @mathhelpforum