Problem 4
Draw any right triangle you want. Afterward, draw a perpendicular segment that goes from the 90 degree angle to the hypotenuse. Call this new segment the altitude. It might help to have the hypotenuse laying flat or horizontal.
The length of this altitude can be found using the geometric mean formula. Search out "geometric mean for triangles" (or similar) and you should get a diagram that visually summarizes what is going on. I've provided a screenshot of an example using GeoGebra. See below.
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Problem 5
The two special types of right triangles are: the 45-45-90 triangle and the 30-60-90 triangle.
If x is the leg length of the first triangle type mentioned, then [tex]x\sqrt{2}[/tex] is the hypotenuse. We can confirm this using the pythagorean theorem. Recall that the legs of any 45-45-90 triangle are the same, meaning we have an isosceles triangle.
For the second type of triangle, the short leg x leads to the hypotenuse 2x and long leg [tex]x\sqrt{3}[/tex]
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Problem 6
If you know say the opposite and adjacent sides of a right triangle, then you can use the tangent ratio because tan = opposite/adjacent.
Sine is used for opposite/hypotenuse, while cosine is used for adjacent/hypotenuse. Those are the main 3 trig functions. Technically there are 3 more, but they are just reciprocals of the previous three.
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Problem 7
Use inverse trig functions to find the missing angles if you know the sides. The inverse trig functions have an exponent of -1.
For instance, let's say the triangle has an opposite side of 10 and hypotenuse of 15
The reference angle would be...
[tex]\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(x) = \frac{10}{15}\\\\x = \sin^{-1}\left(\frac{10}{15}\right)\\\\x \approx 41.81^{\circ}\\\\[/tex]
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Problem 8
Imagine looking completely horizontal at the horizon. The angle in which you are currently looking is 0 degrees. If you look upward, say 10 degrees, then the angle of elevation is 10 degrees. Looking down means we have an angle of depression. These two types of angles help us determine various distances and lengths.
Use the alternate interior angle theorem to help show that angles of depression are congruent to angles of elevation. It will depend on context which type of angle is more useful.
