Recall that if two tangents intersect in the exterior of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs.
See solution.
Practice makes perfect
Let's consider a triangle PQR such that m∠P = 50^(∘) and m∠Q = 60^(∘). Also, we will draw the circle inscribed within △ PQR.
To start, we should recall what Theorem 10.14 states.
Theorem 10.14
If two secants, a secant and a tangent, or two tangents intersect in the exterior of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs.
Notice that the sides of the triangle are all tangent to the circle and intersect outside the circle. Then, we can use Theorem 10.14 to write the following relations.
Tangents
Equation
PA and PB
m ∠P = 1/2(mACB-m AB)
QB and QC
m ∠Q = 1/2(mBAC-m BC)
On the other hand, by the Arc Addition Postulate we have that the sum of the measures of ACB and AB is equal to 360^(∘).
mACB+m AB = 360^(∘)
⇓
mACB = 360^(∘) - m AB
Let's substitute this equation and m ∠P=50^(∘) into the first equation of the table.
Applying the same reasoning, we can write an equation relating mBAC and m BC.
mBAC = 360^(∘) - m BC
Let's substitute this equation and m ∠Q=60^(∘) into the second equation of the table.
Finally, to find the measure of the third arc we use the Arc Addition Postulate again.
m AB_(130^(∘)) + m BC_(120^(∘)) + m AC = 360^(∘)
⇓
m AC = 110^(∘)
We have found the measures of the three arcs.
