If x = 10 m, BC = 80 m, α = 10°, what is β

It seems that I've been recently acquainted with the more difficult parts of Trigonometry. T_T Please help me out?

If x = 10 m, BC = 80 m, α = 10°, what is β?
Here is the image: http://i53.tinypic.com/f3f8te.jpg

tan α = BC / AB, u have α and BC so u get AB
tan (α + β ) = (x + BC) / AB
tan ( 10 + β) = 90 / AB ( which u have)
10 + β = tan ^ (-1) ( 90/AB)
and u get β
hope this was helpful

The way to solve this problem is first solve ΔABC for AB and then solve for β in ΔABD.

α = 10°
BC = 80m
tan(10°) = BC/AB = 80m/AB
AB = 80m x tan(10°) = 14.1m

Now we can solve for β.

tan(α + β) = (BC + x)/AB = 90m/14.1m = 6.38

tan^-1(6.38) (also known as arctan or cotan) = 81.09°.

α + β = 81.09°
81.09° - 10° = 71.09° = β

Again, β = 71.09°

Hope this helps.

line AB = 80/tan(α) by using right triangle tanα = CB/AB
tan(α +β) = (BC + x) / (AB)
α +β = 11.21998593°
β = 11.21998593° - 10° = 1.21998593° ---- answer