Emergency Maths Help

A person on top of a building 33 metres high looks across the road to a church on the other side.

The angle of depression of the bottom of the church is 25 degrees and

the angle of elevation of the top of the church is 18 degrees.

Calculate the height of the church.

Let h = height of the church
Let d = horiz. distance building and church

\(\begin{array}{|rcll|} \hline (1) & \tan(18^{\circ}) &=& \frac{h-33\ m}{d} \\ (2) & \tan(25^{\circ}) &=& \frac{33\ m}{d} \\ \hline & d = \frac{h-33\ m}{\tan(18^{\circ})} &=& \frac{33\ m}{\tan(25^{\circ})} \\ & \frac{h-33}{\tan(18^{\circ})} &=& \frac{33}{\tan(25^{\circ})} \\ & h-33 &=& \frac{ \tan(18^{\circ}) } { \tan(25^{\circ}) } \cdot 33 \\ & h &=& 33 + \frac{ \tan(18^{\circ}) } { \tan(25^{\circ}) } \cdot 33 \\ & h &=& 33\cdot \left(1 + \frac{ \tan(18^{\circ}) } { \tan(25^{\circ}) } \right) \\ & h &=& 33\cdot \left(1 + \frac{ 0.32491969623 } { 0.46630765815 } \right) \\ & h &=& 33\cdot 1.69679253718 \\ & h &=& 55.9941537270 \ m\\ \hline \end{array}\)

The height of the church is \(\mathbf{\approx 56\ m}\)

Interesting enough... No wonder it burnt your brain, here we go:

Illustration (Made om quick session in 3 mins)

We know the height of the building is 33 meters, therefore we obtain the following equation:

Set the distance from the foundation of the building to the church's foundation \(A\).\

Apply trigonometric identity.

The angle of  depression is 25 degrees, therefore:

\(A=tan(25°)*33\)

\(A≈ 0.4663*25\)

\(A≈ 11.6575\)

That means the distance from the church's top to the same height of the building is also \(A=11.6575\)

Applying trigonometric identites for the second time

Church height=\(11.6575*sec(18°)\)

\(≈11.6575*1.0515\)

\(=12.258\)=Church's Height

Q.E.D.

(Don't even ask me about how tensed I am to win over Heureka over the speed of solving this question)