Question

There is a tower the angle of elevation from the point 'A' which is situated South of the tower is 30°. And angle of elevation of the tower from point 'B' which is to the West of point 'A' is 18°.
If AB = a
Find the height of tower using 'a'.
(sin 18°= (√5-1)/4 )

Answer 1

Answer on Question #65904 – Math – Trigonometry

Question

There is a tower the angle of elevation from the point 'A' which is situated South of the tower is $30{}^{\circ}$. And angle of elevation of the tower from point 'B' which is to the West of point 'A' is $18{}^{\circ}$.

If $AB = a$

Find the height of tower using 'a'. (sin $18{}^{\circ} = (v5 - 1)/4$).

Solution

Consider these pictures:

Let $h$ be the height of tower, $|AT| = b$, $|BT| = c$.

According to Pythagorean theorem, $a^2 + b^2 = c^2$.

At the same time $\cot 30{}^{\circ} = \frac{b}{h}$ and $\cot 18{}^{\circ} = \frac{c}{h}$. Therefore $b = h \cot 30{}^{\circ}$ and $c = h \cot 18{}^{\circ}$.

$$a^2 + (h \cot 30{}^{\circ})^2 = (h \cot 18{}^{\circ})^2 \Rightarrow a^2 = h^2 (\cot^2 18{}^{\circ} - \cot^2 30{}^{\circ}) \Rightarrow$$ $$h = \frac{a}{\sqrt{\cot^2 18{}^{\circ} - \cot^2 30{}^{\circ}}} = \frac{a}{\sqrt{\cot^2 18{}^{\circ} - (\sqrt{3})^2}}$$ $$\cot^2 18{}^{\circ} = \frac{1}{(\sin 18{}^{\circ})^2} - 1 = \frac{1}{\left(\frac{\sqrt{5} - 1}{4}\right)^2} - 1 = \frac{16}{(\sqrt{5} - 1)^2} - 1 = \frac{16 - (\sqrt{5} - 1)^2}{(\sqrt{5} - 1)^2} = \frac{10 + 2\sqrt{5}}{(\sqrt{5} - 1)^2}$$ $$h = \frac{a}{\sqrt{\frac{10 + 2\sqrt{5}}{(\sqrt{5} - 1)^2} - 3}} = \frac{a(\sqrt{5} - 1)}{\sqrt{10 + 2\sqrt{5} - 3(\sqrt{5} - 1)^2}} = \frac{a(\sqrt{5} - 1)}{\sqrt{8(\sqrt{5} - 1)}} = \frac{\sqrt{\sqrt{5} - 1}}{2\sqrt{2}} a$$

Answer: $\frac{\sqrt{\sqrt{5} - 1}}{2\sqrt{2}} a$

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