# Math Grade 11 Sine and Cosine lawsThe Leaning Tower of Pisa leans toward the south at an angle of 5.5°. One day near noon its shadow was measured to be 84.02 m long and the angle of elevation from...

Math Grade 11 Sine and Cosine laws

The Leaning Tower of Pisa leans toward the south at an angle of 5.5°. One day near noon its shadow was measured to be 84.02 m long and the angle of elevation from the tip of the shadow to the top of the tower at that time was measured as 32.0°. To answer the following, assume that the Tower is like a pole stuck in the ground, that is, it has negligible width. (Also it is important to know that Pisa Italy is at a latitude of approximately 44°N because this affects the direction of the shadow!)

- Determine the slant height of the tower.
- How high is the tip of the tower above the ground?

please draw a diagram out so i can understand easily.

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Since we do not know the exact date or time we assume that the shadow falls due north. (Pisa is north of the Tropic of Cancer so the sun will appear in the southern sky around noon.)

The tower leans `5.5^@` from the vertical towards the south. The length of the shadow to the north is 84.02m and the angle of elevation is `32.0^@` . Draw a triangle with side 84.02 such that this side is included between the angles measuring `32^@` and `95.5^@` . (We use `95.5^@` since the tower if vertical would form a 90 degree angle; but it leans `5.5^@` away from the shadow.) The third angle of the triangle, opposite the side of known length, will be `52.5^@` .

Let the slant height be `l` and the vertical height be h.

(a) From the Law of Sines we have `(sin52.5^@)/84.02=(sin32^@)/l`

Then `l=(84.02sin32^@)/(sin52.5^@)~~56.12`

**So the slant height is approximately 56.12m**

(b) The vertical height can be found by dropping a perpendicular from the vertex of the angle with measure `52.5^@` to the extended side opposite. This forms a right triangle with hypotenuse approximately 56.12m and the angle opposite the vertical leg `84.5^@` .

Then the vertical leg has a length found by:

`sin85.5^@=h/56.12`

So `h=56.12sin84.5^@~~55.86m`

**Then the vertical height is approximately 55.86m.**

The bottom of the triangle is heading north from right to left; the red perpendicular is dropped from the south side. The angle to the far left is 32 degrees; the angle at the top of the triangle is 52.5 degrees.