Originally Posted by

**Grandad** Hello thereddevilsHave a look at the sketch I've attached.

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

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

The distance AB = $\displaystyle l$.

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

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

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

You can do it!

Grandad