Trignometry - modelling and problem solving

• May 12th 2013, 08:31 PM
elmidge
Trignometry - modelling and problem solving
From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

Please can you help with this problem
• May 18th 2013, 01:08 PM
x3bnm
Re: Trignometry - modelling and problem solving
Quote:

From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

Please can you help with this problem
Attachment 28407

From the image above Given: $\angle EAD = 45^\circ$, $\angle BAD = 120^\circ$ and $AB = 100m$

We have to find the height of the tower $ED$

Draw from point $G$ to $CD$, $GH$ which is perpendicular to $CD$ and to $BE$, $GO$ perpendicular to $BE$

In $\triangle ABC$:
\begin{align*}\sin {\angle BAC} =& \frac{BC}{AB}\\ =& \frac{BC}{100}....\text{[because }AB = 100, \angle BAD = 120^\circ \text{ and }\angle BAC = 180^\circ - 120^\circ = 60^\circ\text{]}\end{align*}
$\therefore GH = BC = \frac{100\sqrt{3}}{2}$

In $\triangle GAH$:
$\sin {\angle GAH} = \frac{GH}{AG}$
$\therefore AG = \frac{100\sqrt{3}}{\sqrt{2}}....\text{[} \angle GAH = 45^\circ \text{ and } GH = \frac{100\sqrt{3}}{2}}\text{]}$

In $\triangle ABG$:
\begin{align*}BG =& \sqrt{AB^2 + AG^2 - 2*AB*AG*\cos {\angle BAG}}...\text{[ by cosine law]}\\=& 136.60...[AB = 100, AG = \frac{100\sqrt{3}}{\sqrt{2}}, \angle BAG = 75^\circ]\end{align*}

In $\triangle BOG$:
$\sin{\angle OBG} = \frac{OG}{BG}$
$\therefore OG = 59.88^\circ...[\angle OBG = 26^\circ \text{ and } BG = 136.60$

In $\triangle BEF$:
\begin{align*}\angle FEB =& 180^\circ - \angle FBE - \angle BFE \\ =& 180^\circ - 26^\circ - 90^\circ\\=& 64^\circ\end{align*}
Now $\angle AGH = \angle GEF$ because $GF$ is parallel to $AD$ and straight line $AE$ intersects them.

In $\triangle AGH$:
$\angle AGH = 180^\circ - \angle GAH - \angle GHA = 45^\circ...[\angle GAH = 45^\circ, \angle GHA = 90^\circ$

Now:
$\angle FEB = \angle BEA + \angle AED$

Because $\angle AED = \angle GEF = \angle AGH = 45^\circ\text{ and } \angle FEB = 64^\circ$:

$\therefore \angle BEA = 19^\circ$

In $\triangle OEG$:
$\sin{\angle GEO} = \frac{OG}{GE}$

Because $\angle BEA = 19^\circ \text{ and } OG = 59.88$
$\therefore GE = 183.92$

In $\triangle GEF$:
$\cos{\angle GEF} = \frac{EF}{GE}$

But $\angle GEF = 45^\circ \text{ and } GE = 183.92$
$\therefore EF = 130.05$

Now:
\begin{align*}ED = EF + FD = EF + BC =& 130.05 + \frac{100\sqrt{3}}{2}\\ =& 216.65\end{align*}

So the height of the tower is $216.65 \text{ meter}$