1. ## 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

$
\frac{l\tan\alpha\sin\omega\cos\theta}{\sin(\omega-\phi)}
$

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

$
\tan\beta =\csc \phi[\tan\alpha\sin\omega-\tan\theta\sin(\omega-\phi)]
$

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 .

2. ## 3D Trigonometry

Hello thereddevils
Originally Posted by thereddevils
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

$
\frac{l\tan\alpha\sin\omega\cos\theta}{\sin(\omega-\phi)}
$

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

$
\tan\beta =\csc \phi[\tan\alpha\sin\omega-\tan\theta\sin(\omega-\phi)]
$

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!

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!