1. triangle problem solving questions.

1)From the window of one building, Matt finds the angle of elevation to the top of a second building to be 33degrees and the angle of depression to the bottom of the same building to be 55degrees. The buildings are 25m apart. Find the height of the second building.

2)Two ships leave the same harbour at the same time, one at a speed of 12km/hr on a course of 20degrees, the other at a speed of 15km/hr on a course of 254degrees. How far apart will the ships be after 4 hours.

If you can do one go ahead and do it.

2. [QUOTE=jimC;168457]1)From the window of one building, Matt finds the angle of elevation to the top of a second building to be 33degrees and the angle of depression to the bottom of the same building to be 55degrees. The buildings are 25m apart. Find the height of the second building.
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1) Make a sketch. Draw a horizontal line from Matt's eye to the next building. As you easily can see you are dealing with 2 right triangles.

Use the tan-function:

$\displaystyle \tan(33^\circ) = \frac x{25}\ \text{ and }\ \tan(55^\circ) = \frac y{25}$

Then the height of the building is $\displaystyle h = x+y$

For your confirmation only: $\displaystyle h = 51.94\ m$

3. Originally Posted by jimC
...

2)Two ships leave the same harbour at the same time, one at a speed of 12km/hr on a course of 20degrees, the other at a speed of 15km/hr on a course of 254degrees. How far apart will the ships be after 4 hours.

...
The 2 courses and distances are the sides of a triangle:

$\displaystyle s_1 = 12\ \frac{km}h \cdot 4 \ h = 48\ km$

$\displaystyle s_2 = 15\ \frac{km}h \cdot 4 \ h = 60\ km$

They include an angle of

$\displaystyle \alpha = 254^\circ - 20^\circ = 234^\circ$

Use Cosine rule to calculate the third side which represents the distance between the ships.

For your confirmation only : $\displaystyle s_3 \approx 96.383\ km$