Question #023a8

1 Answer
Nov 28, 2015

(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.