# Question #023a8

Nov 28, 2015

(i) $A B = 25.7$ m
(ii) ${38.7}^{\circ}$
(iii) $D B = 29.8$ m
(iv) ${21.9}^{\circ}$

#### 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:
$\tan \left(25\right) = \frac{B T}{A B}$
so
$A B = \frac{B T}{\tan \left(25\right)} = \frac{12}{\tan \left(25\right)}$
which is
$A B = 25.734 \ldots \approx \textcolor{b l u e}{25.7}$ m

(ii) Consider the triangle CBT.
Same relation as above. Let x be the angle $\angle B C T$.
Then:
$\tan \left(x\right) = \frac{B T}{B C} = \frac{12}{15}$
so,
$x = {\tan}^{-} 1 \left(\frac{12}{15}\right) = 38.6598 \ldots \approx {\textcolor{b l u e}{38.7}}^{\circ}$

(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:
$D {B}^{2} = A {B}^{2} + A {D}^{2}$
or
$D {B}^{2} = A {B}^{2} + B {C}^{2}$ since AD=BC
i.e.
$D B = \sqrt{{\left(\frac{B T}{\tan \left(25\right)}\right)}^{2} + {15}^{2}}$
i.e.
$D B = 29.786 \ldots \approx \textcolor{b l u e}{29.8}$ m

(iv) Consider the triangle DBT.
Let $\alpha$ be the angle $\angle B D T$.
Then:
$\tan \left(\alpha\right) = \frac{B T}{D B} = \frac{12}{\sqrt{{\left(\frac{B T}{\tan \left(25\right)}\right)}^{2} + {15}^{2}}}$
or
$\tan \left(\alpha\right) = \frac{12}{29.8}$
so
$\alpha = {\tan}^{-} 1 \left(\frac{12}{29.8}\right)$
i.e.
$\alpha = 21.9427 \ldots \approx {\textcolor{b l u e}{21.9}}^{\circ}$

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 ${38.7}^{\circ}$
25.7 m (AB), the angle is ${25}^{\circ}$
29.8 m (DB), the angle is ${21.9}^{\circ}$

I hope these help.