Solve Right Triangle Problems

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

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.

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.
   
   
   
   
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).

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.

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)]

• Triangle Problems
  
  
• Sine Law - Ambiguous case - applet