Exercise 22 Page 654

In the given diagram, we want to find m BC, which is a minor arc.

Note that the major arc BTC and the minor arc BC are adjacent arcs. By the Arc Addition Postulate, we know that the measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs. In this case, the arc formed by these two adjacent arcs is a full circle, which measures 360^(∘). m BTC+m BC=360^(∘)

To find the measure of BC, we have to find first m BTC. By looking at the diagram, we note that BTC is formed by the adjacent arcs BT and TC.

We can see in the diagram that the central angle for BT is a right angle. Recall that the measure of a minor arc is equal to the measure of its corresponding central angle. mBT=m∠ BPT ⇔ mBT=90^(∘) In the same way, we see that the central angle for the minor arc TC is 128^(∘). By the Arc Addition Postulate, we know that the measure of the major arc BTC is the sum of the measures of BT and TC. m BTC=mBT+mTC Let's substitute 90^(∘) for mBT and 128^(∘) for mTC in the above equation, and solve for BTC.

mBTC=mBT+mTC

SubstituteII

mBT= 90, mTC= 128

mBTC=90+128

AddTerms

Add terms

mBTC=218

Therefore, the measure of BTC is 218^(∘). Now, we will substitute 218^(∘) for m BTC into the first equation, and solve for m BC.

mBTC+mBC=360

Substitute

mBTC= 218

218+mBC=360

SubEqn

LHS-218=RHS-218

mBC=142

We found that BC measures 142^(∘).