Solve a right triangle whose sides are all tangent to a circle. Both the problem and its detailed solution are presented.
Problem Statement
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.
Step-by-Step Solution
Step 1: Analyzing Angles via Congruent Triangles
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. The size of angle OCM is given by:
\[ \angle OCM = \frac{36^\circ}{2} = 18^\circ \]Step 2: Calculating Length CM
Using \(\tan(\angle OCM)\) to calculate the length of side CM:
\[ \tan(\angle OCM) = \tan(18^\circ) = \frac{r}{CM} \] \[ CM = \frac{r}{\tan(18^\circ)} = \frac{10}{\tan(18^\circ)} \approx 30.8 \text{ cm} \]Step 3: Calculating Angle \(\angle PAM\)
The size of angle \(\angle PAM\) is:
\[ 90^\circ - 36^\circ = 54^\circ \]Step 4: Finding Angle \(\angle OAM\)
Since triangles AOM and AOP are right and congruent, the size of angle \(\angle OAM\) is:
\[ \frac{54^\circ}{2} = 27^\circ \]Step 5: Calculating Length AM
The length of side AM is calculated as:
\[ \tan(\angle OAM) = \tan(27^\circ) = \frac{r}{AM} \] \[ AM = \frac{r}{\tan(27^\circ)} = \frac{10}{\tan(27^\circ)} \approx 19.6 \text{ cm} \]Step 6: Finding the Hypotenuse AC
The hypotenuse AC is:
\[ AC = AM + CM = 19.6 + 30.8 = 50.4 \text{ cm} \]Step 7: Finding Side AB
The side AB is:
\[ AB = AP + r = AM + r = 19.6 + 10 = 29.6 \text{ cm} \]Step 8: Finding Side BC
The side BC is:
\[ BC = BN + CN = 10 + CM = 10 + 30.8 = 40.8 \text{ cm} \]