Question

Question #023a8

Answer 1

(i) `AB=25.7` m
(ii) `38.7^@`
(iii) `DB=29.8` m
(iv) `21.9^@`

Explanation:

I assume that I can use a calculator for this because 25 degrees is not really a great number to work with in trigonometry. I will also ignore the [ ] (brackets) and | | (pipes) notations.
Okay, let's do this.

(i) Consider the triangle ABT.
We know the following relation:
`tan(25)=(BT)/(AB)`
so
`AB=(BT)/(tan(25))=12/(tan(25))`
which is
`AB=25.734... ~~color(blue)25.7` m

(ii) Consider the triangle CBT.
Same relation as above. Let x be the angle `/_BCT`.
Then:
`tan(x)=(BT)/(BC)=12/15`
so,
`x=tan^-1(12/15)=38.6598...~~color(blue)38.7^@`

(iii) Consider the triangle ABD (or BCD).
Use the Pythagorean theorem (I'm the one who found it, I'm such a genious) to get DB:
`DB^2=AB^2+AD^2`
or
`DB^2=AB^2+BC^2` since AD=BC
i.e.
`DB=sqrt(((BT)/(tan(25)))^2 + 15^2)`
i.e.
`DB=29.786...~~color(blue)29.8` m

(iv) Consider the triangle DBT.
Let `alpha` be the angle `/_BDT`.
Then:
`tan(alpha)=(BT)/(DB)=12/sqrt(((BT)/(tan(25)))^2 + 15^2)`
or
`tan(alpha)=12/29.8`
so
`alpha=tan^-1(12/29.8)`
i.e.
`alpha=21.9427...~~color(blue)21.9^@`

That's it!
Just to have a little bit more confidence in our answer, and since the mast has a fixed height (12 m) you should notice that the longer the baseline is, the smaller should the angle be.
When the baseline is:

15 m (BC), the angle is `38.7^@`
25.7 m (AB), the angle is `25^@`
29.8 m (DB), the angle is `21.9^@`

I hope these help.