Circle Tangent to Right Triangle - Problem With Solution

Solve a right triangle whose sides are all tangent to a circle. Both the problem and its detailed solution are presented.

Problem

ABC is a right triangle. Given one of the angles of triangle ABC and the radius of the circle tangent to all three sides of the right triangle, find the lengths of the two sides and the hypotenuse of triangle ABC.

Solution

• Triangles COM and CON are right triangles with congruent hypotenuses \( OC \) and congruent corresponding sides \( OM \) and \( ON \), so they are congruent. Hence, angles \( \angle OCM \) and \( \angle OCN \) are also congruent; hence the size of angle OCM is given by
  

  

  \[ \angle OCM = \frac{36^\circ}{2} = 18^\circ \]
• Using \( \tan(\angle OCM) \) to calculate the length of side CM: \[ \tan \angle OCM = \tan(18^\circ) = \frac{r}{CM} \] \[ CM = \frac{r}{\tan(18^\circ)} = \frac{10}{\tan(18^\circ)} \approx 30.8 \text{ cm} \]
• The size of angle \( \angle PAM \) is: \[ 90^\circ - 36^\circ = 54^\circ \]
• Since triangles AOM and AOP are right and congruent, the size of angle \( \angle OAM \) is: \[ \frac{54^\circ}{2} = 27^\circ \]
• The length of side AM is calculated as: \[ \tan \angle OAM = \tan(27^\circ) = \frac{r}{AM} \] \[ AM = \frac{r}{\tan(27^\circ)} = \frac{10}{\tan(27^\circ)} \approx 19.6 \text{ cm} \]
• The hypotenuse AC is: \[ AC = AM + CM = 19.6 + 30.8 = 50.4 \text{ cm} \]
• The side AB is: \[ AB = AP + r = AM + r = 19.6 + 10 = 29.6 \text{ cm} \]
• The side BC is: \[ BC = BN + CN = 10 + CM = 10 + 30.8 = 40.8 \text{ cm} \]