Math  /  Geometry

Question7. Suppose Romeo is serenadib facing north and sees the Juliet while she is on her balcony. Romeo is other suitor, is observing balcony at an angle of elevation of 2020^{\circ}. Paris, Juliet's balcony at an angle of the situation and is facing west. Paris sees the shown. Determine of elevation of 1818^{\circ}. Romeo and Paris are 100 m apart as nearest metre.

Studdy Solution

STEP 1

What is this asking? How high up is Juliet's balcony if Romeo and Paris are looking at it from different angles and are 100m apart? Watch out! Don't mix up the angles!
Make sure you're using the right angle with the right distance.
Also, remember we're dealing with a 3D situation, not just a flat triangle.

STEP 2

1. Set up the triangles
2. Find distances from Romeo and Paris to the building
3. Use Pythagorean theorem

STEP 3

Imagine Romeo and Paris looking at Juliet.
They're forming two right-angled triangles: one with Romeo, Juliet, and the base of the building, and another with Paris, Juliet, and the base of the building.
These triangles share a side: the height of the balcony!

STEP 4

Let's call the height of the balcony hh.
The distance from Romeo to the base of the building is rr, and the distance from Paris to the base of the building is pp.

STEP 5

We know Romeo's angle of elevation is 20\textbf{20}^{\circ}.
We can use the tangent function: tan(20)=hr\tan(20^{\circ}) = \frac{h}{r}.
This means r=htan(20)r = \frac{h}{\tan(20^{\circ})}.

STEP 6

Paris's angle of elevation is 18\textbf{18}^{\circ}.
Using the tangent function again: tan(18)=hp\tan(18^{\circ}) = \frac{h}{p}.
This gives us p=htan(18)p = \frac{h}{\tan(18^{\circ})}.

STEP 7

Now, imagine a triangle on the ground formed by Romeo, Paris, and the base of the building.
It's a right-angled triangle, and we know the distance between Romeo and Paris is 100 m\textbf{100} \text{ m}.

STEP 8

We can use the Pythagorean theorem: r2+p2=1002r^2 + p^2 = 100^2.

STEP 9

Substituting the expressions for rr and pp from the tangent calculations, we get: (htan(20))2+(htan(18))2=1002 \left(\frac{h}{\tan(20^{\circ})}\right)^2 + \left(\frac{h}{\tan(18^{\circ})}\right)^2 = 100^2 h2tan2(20)+h2tan2(18)=10000 \frac{h^2}{\tan^2(20^{\circ})} + \frac{h^2}{\tan^2(18^{\circ})} = 10000 h2(1tan2(20)+1tan2(18))=10000 h^2 \left(\frac{1}{\tan^2(20^{\circ})} + \frac{1}{\tan^2(18^{\circ})}\right) = 10000 h2=100001tan2(20)+1tan2(18) h^2 = \frac{10000}{\frac{1}{\tan^2(20^{\circ})} + \frac{1}{\tan^2(18^{\circ})}} h=100001tan2(20)+1tan2(18) h = \sqrt{\frac{10000}{\frac{1}{\tan^2(20^{\circ})} + \frac{1}{\tan^2(18^{\circ})}}}

STEP 10

tan(20)0.3640\tan(20^{\circ}) \approx 0.3640 and tan(18)0.3249\tan(18^{\circ}) \approx 0.3249. h1000010.36402+10.32492 h \approx \sqrt{\frac{10000}{\frac{1}{0.3640^2} + \frac{1}{0.3249^2}}} h100007.5244+9.4366 h \approx \sqrt{\frac{10000}{7.5244 + 9.4366}} h1000016.9610 h \approx \sqrt{\frac{10000}{16.9610}} h589.55 h \approx \sqrt{589.55} h24.28 h \approx 24.28

STEP 11

Juliet's balcony is approximately 24\textbf{24} meters high.

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