Solve Right Triangle Problems

This is a tutorial on right triangles problems. Examples with detailed solutions and explanations are included.

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

right triangle

Solution to Problem 1:

  1. First use the Pythagorean theorem to find the hypotenuse h of the right triangle.

    h = SQRT(62 + 82)

    = SQRT(36 + 64)

    = 10

  2. In a right triangle, sin(x) is given by.

    sin(x) = 8 / 10 = 0.8

  3. 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.

lines tangent to circle (1)

Solution to Problem 2:

  1. 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.

    lines tangent to circle (2)

  2. 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.

  3. solve for d(NA)

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

  4. Use the Pythagorean theorem in the right triangle NAC to find the distance d(AC)

    d(AC) = SQRT[ r2 + d(NA)2]

  5. Substitute d(NA) by r / tan(x/2) to obtain d(AC).

    d(AC) = SQRT[ r2 + (r / tan(x/2))2]

  6. Factor r2

    d(AC) = SQRT[ r2(1 + 1 / tan(x/2)2]

  7. 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).

right triangle with common side

Solution to Problem 3:

  1. We first use the right triangle on the right to find a
    tan(410) = a / 15

  2. a is given by.
    a = 15 * tan(410)

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

  4. Substitute a by 15 * tan(410) in the above.
    tan(x) = 15 * tan(410) / 10

  5. Which gives
    tan(x) = 1.5 * tan(410)

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.

tower viewed from two points

Solution to Problem 4:

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

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

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

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

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

  6. Solve the above for h to obtain.
    h = d tan(a) tan(b) / [ tan(b) - tan(a)]

More references on solving triangle problems.

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Updated: February 2015

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