How to find the height of the mountain

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The angle of elevation on the top of a mountain from a point on the ground is found to be $alpha$. After walking a distance of $a$ along a slope of inclination $beta$ towards the cliff, the angle of elevation is found to be $gamma$. Show that the height of the mountain is
$$fraca sin alpha sin (alpha-beta)sin(gamma-alpha)$$ .



I tried it this way:
My drawing



From here, I calculated $BE$ to be
$$fracasin(alpha-beta)sin(gamma-alpha)$$
Also, I got $FE$ to be $asin(alpha-beta)$.
Height $AO$ is $asinalpha$ from the image.



I am unable to proceed hereafter. Any help would be appreciated. Thanks in advance.




trigonometry




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  • I think you are confused with x and y planes when he is walking at the $beta$ angle he is walking in the x plane whereas the mountain is in y plane
    – Deepesh Meena
    Aug 22 at 11:09











  • Do you mean that the angle of elevation is found to be $gamma$?
    – N. F. Taussig
    Aug 23 at 7:09














up vote
0
down vote

favorite












The angle of elevation on the top of a mountain from a point on the ground is found to be $alpha$. After walking a distance of $a$ along a slope of inclination $beta$ towards the cliff, the angle of elevation is found to be $gamma$. Show that the height of the mountain is
$$fraca sin alpha sin (alpha-beta)sin(gamma-alpha)$$ .



I tried it this way:
My drawing



From here, I calculated $BE$ to be
$$fracasin(alpha-beta)sin(gamma-alpha)$$
Also, I got $FE$ to be $asin(alpha-beta)$.
Height $AO$ is $asinalpha$ from the image.



I am unable to proceed hereafter. Any help would be appreciated. Thanks in advance.




trigonometry




share|cite|improve this question






















  • I think you are confused with x and y planes when he is walking at the $beta$ angle he is walking in the x plane whereas the mountain is in y plane
    – Deepesh Meena
    Aug 22 at 11:09











  • Do you mean that the angle of elevation is found to be $gamma$?
    – N. F. Taussig
    Aug 23 at 7:09












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The angle of elevation on the top of a mountain from a point on the ground is found to be $alpha$. After walking a distance of $a$ along a slope of inclination $beta$ towards the cliff, the angle of elevation is found to be $gamma$. Show that the height of the mountain is
$$fraca sin alpha sin (alpha-beta)sin(gamma-alpha)$$ .



I tried it this way:
My drawing



From here, I calculated $BE$ to be
$$fracasin(alpha-beta)sin(gamma-alpha)$$
Also, I got $FE$ to be $asin(alpha-beta)$.
Height $AO$ is $asinalpha$ from the image.



I am unable to proceed hereafter. Any help would be appreciated. Thanks in advance.




trigonometry




share|cite|improve this question














The angle of elevation on the top of a mountain from a point on the ground is found to be $alpha$. After walking a distance of $a$ along a slope of inclination $beta$ towards the cliff, the angle of elevation is found to be $gamma$. Show that the height of the mountain is
$$fraca sin alpha sin (alpha-beta)sin(gamma-alpha)$$ .



I tried it this way:
My drawing



From here, I calculated $BE$ to be
$$fracasin(alpha-beta)sin(gamma-alpha)$$
Also, I got $FE$ to be $asin(alpha-beta)$.
Height $AO$ is $asinalpha$ from the image.



I am unable to proceed hereafter. Any help would be appreciated. Thanks in advance.





trigonometry





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 at 23:53









N. F. Taussig

38.8k93153




38.8k93153










asked Aug 22 at 10:49









Chrys

24




24











  • I think you are confused with x and y planes when he is walking at the $beta$ angle he is walking in the x plane whereas the mountain is in y plane
    – Deepesh Meena
    Aug 22 at 11:09











  • Do you mean that the angle of elevation is found to be $gamma$?
    – N. F. Taussig
    Aug 23 at 7:09
















  • I think you are confused with x and y planes when he is walking at the $beta$ angle he is walking in the x plane whereas the mountain is in y plane
    – Deepesh Meena
    Aug 22 at 11:09











  • Do you mean that the angle of elevation is found to be $gamma$?
    – N. F. Taussig
    Aug 23 at 7:09















I think you are confused with x and y planes when he is walking at the $beta$ angle he is walking in the x plane whereas the mountain is in y plane
– Deepesh Meena
Aug 22 at 11:09





I think you are confused with x and y planes when he is walking at the $beta$ angle he is walking in the x plane whereas the mountain is in y plane
– Deepesh Meena
Aug 22 at 11:09













Do you mean that the angle of elevation is found to be $gamma$?
– N. F. Taussig
Aug 23 at 7:09




Do you mean that the angle of elevation is found to be $gamma$?
– N. F. Taussig
Aug 23 at 7:09










1 Answer
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The formula for the height is wrong.



enter image description here



We know $angleBCD=fracpi2-gamma$ and $angleBCA=fracpi2-alpha$ so therefore $angleDCA=fracpi2-alpha - (fracpi2-gamma)=gamma-alpha$. We also know $angleDAC=alpha-beta$. We therefore know that $angle ADC=pi +beta-gamma$. Using the law of sines we can find that $$fracsin (angleADC)AC = fracsin(angleDCA)a$$
or
$$fracsin (pi+beta-gamma)AC = fracsin(gamma-alpha)a$$
or
$$AC= fracasin(gamma-beta)sin(gamma-alpha)$$



As the height of the mountain is $h=ACsin(alpha)$, we arrive at the formula $$h =fracasin(alpha)sin(gamma-beta)sin(gamma-alpha)$$






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    1 Answer
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    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    The formula for the height is wrong.



    enter image description here



    We know $angleBCD=fracpi2-gamma$ and $angleBCA=fracpi2-alpha$ so therefore $angleDCA=fracpi2-alpha - (fracpi2-gamma)=gamma-alpha$. We also know $angleDAC=alpha-beta$. We therefore know that $angle ADC=pi +beta-gamma$. Using the law of sines we can find that $$fracsin (angleADC)AC = fracsin(angleDCA)a$$
    or
    $$fracsin (pi+beta-gamma)AC = fracsin(gamma-alpha)a$$
    or
    $$AC= fracasin(gamma-beta)sin(gamma-alpha)$$



    As the height of the mountain is $h=ACsin(alpha)$, we arrive at the formula $$h =fracasin(alpha)sin(gamma-beta)sin(gamma-alpha)$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      The formula for the height is wrong.



      enter image description here



      We know $angleBCD=fracpi2-gamma$ and $angleBCA=fracpi2-alpha$ so therefore $angleDCA=fracpi2-alpha - (fracpi2-gamma)=gamma-alpha$. We also know $angleDAC=alpha-beta$. We therefore know that $angle ADC=pi +beta-gamma$. Using the law of sines we can find that $$fracsin (angleADC)AC = fracsin(angleDCA)a$$
      or
      $$fracsin (pi+beta-gamma)AC = fracsin(gamma-alpha)a$$
      or
      $$AC= fracasin(gamma-beta)sin(gamma-alpha)$$



      As the height of the mountain is $h=ACsin(alpha)$, we arrive at the formula $$h =fracasin(alpha)sin(gamma-beta)sin(gamma-alpha)$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        The formula for the height is wrong.



        enter image description here



        We know $angleBCD=fracpi2-gamma$ and $angleBCA=fracpi2-alpha$ so therefore $angleDCA=fracpi2-alpha - (fracpi2-gamma)=gamma-alpha$. We also know $angleDAC=alpha-beta$. We therefore know that $angle ADC=pi +beta-gamma$. Using the law of sines we can find that $$fracsin (angleADC)AC = fracsin(angleDCA)a$$
        or
        $$fracsin (pi+beta-gamma)AC = fracsin(gamma-alpha)a$$
        or
        $$AC= fracasin(gamma-beta)sin(gamma-alpha)$$



        As the height of the mountain is $h=ACsin(alpha)$, we arrive at the formula $$h =fracasin(alpha)sin(gamma-beta)sin(gamma-alpha)$$






        share|cite|improve this answer












        The formula for the height is wrong.



        enter image description here



        We know $angleBCD=fracpi2-gamma$ and $angleBCA=fracpi2-alpha$ so therefore $angleDCA=fracpi2-alpha - (fracpi2-gamma)=gamma-alpha$. We also know $angleDAC=alpha-beta$. We therefore know that $angle ADC=pi +beta-gamma$. Using the law of sines we can find that $$fracsin (angleADC)AC = fracsin(angleDCA)a$$
        or
        $$fracsin (pi+beta-gamma)AC = fracsin(gamma-alpha)a$$
        or
        $$AC= fracasin(gamma-beta)sin(gamma-alpha)$$



        As the height of the mountain is $h=ACsin(alpha)$, we arrive at the formula $$h =fracasin(alpha)sin(gamma-beta)sin(gamma-alpha)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 22 at 17:03









        Jens

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