Example  Problem 1: Find sin(x) and cos(x) in the right triangle shown below.
Solution to Problem 1:

First use the Pythagorean theorem to find the hypotenuse h of the right triangle.
h = SQRT(6^{2} + 8^{2})
= SQRT(36 + 64)
= 10

In a right triangle, sin(x) is given by.
sin(x) = 8 / 10 = 0.8

and cos(x) is given by
cos(x) = 6 / 10 = 0.6
Example  Problem 2: Two lines tangent to a circle at points M and N have a point of intersection A. The size of angle MAN is equal to x degrees and the length of the radius of the circle is equal to r. Find the distance from point A to the center of the circle in terms of x and r.
Solution to Problem 2:

A line through the center C of the circle and a point of tangency to the circle is perpendicular to the tangent line, hence the right angles at M and N in the figure below.

Tringles MAC and NAC are congruent right triangles. tan(x/2) is given by.
tan(x/2) = r / d(NA) , where d(NA) = distance between points A and N.

solve for d(NA)
d(NA) = r / tan(x/2)

Use the Pythagorean theorem in the right triangle NAC to find the distance d(AC)
d(AC) = SQRT[ r^{2} + d(NA)^{2}]

Substitute d(NA) by r / tan(x/2) to obtain d(AC).
d(AC) = SQRT[ r^{2} + (r / tan(x/2))^{2}]

Factor r^{2}
d(AC) = SQRT[ r^{2}(1 + 1 / tan(x/2)^{2}]

Simplify
d(AC) = r SQRT[ 1 + 1 / tan(x/2)^{2}]
Example  Problem 3: Two right triangles have side a in common. x is the size of angle BAC. Find tan(x).
Solution to Problem 3:

We first use the right triangle on the right to find a
tan(41^{0}) = a / 15

a is given by.
a = 15 * tan(41^{0})

We now use the right triangle on the left to find tan(x).
tan(x) = a / 10

Substitute a by 15 * tan(41^{0})
in the above.
tan(x) = 15 * tan(41^{0}) / 10

Which gives
tan(x) = 1.5 * tan(41^{0})
Example  Problem 4: From point A, an observer notes that the angle of elevation of the top of a tower (C,D) is a (degrees) and from point B the angle of elevation is b (degrees). Points A, B and C (the bottom of the tower) are collinear. The distance between A and B is d. Find the height h of the tower in terms of d and angles a and b.
Solution to Problem 4:

Let x be the distance between points B and C, hence in the right triangle ACD we have
tan(a) = h / (d + x)

and in the right triangle BCD we have
tan(b) = h / x

Solve the above for x
x = h / tan(b)

Solve tan(a) = h / (d + x) for h
h = (d + x) tan(a)

Substitute x in above by h / tan(b)
h = (d + h / tan(b)) tan(a)

Solve the above for h to obtain.
h = d tan(a) tan(b) / [ tan(b)  tan(a)]
More references on solving triangle problems.