Exercise 37 Page 766

Let's begin by marking some points in the given diagram.

To find x we first have to find m∠ A. We can do this by noticing that AB and AC are secant to the smaller circle and they intersect each other outside it. Thus, the measure of ∠ A is one half the measure of the difference of the intercepted arcs. m∠ A = 1/2(m FG - m HI) Next, let's substitute the given measures into the equation above.

m∠ A = 1/2(m FG - m HI)

SubstituteII

m FG= 118^(∘), m HI= 54^(∘)

m∠ A = 1/2( 118^(∘) - 54^(∘))

▼

Simplify

SubTerms

Subtract terms

m∠ A = 1/2(64^(∘))

Multiply

Multiply

m∠ A = 64^(∘)/2

CalcQuot

Calculate quotient

m∠ A = 32^(∘)

Now that we know the measure of ∠ A we apply the same reasoning as before, but this time we will use the bigger circle. Thus, the measure of ∠ A equals one half the measure of the difference of the intercepted arcs. m∠ A = 1/2(m BC - m DE_(x^(∘))) Finally, we substitute the corresponding values and find the value of x.

m∠ A = 1/2(m BC - x^(∘))

SubstituteII

m BC= 79^(∘), m∠ A= 32^(∘)

32^(∘) = 1/2( 79^(∘) - x^(∘))

▼

Solve for x^(∘)

MultEqn

LHS * 2=RHS* 2

64^(∘) = 79^(∘) - x^(∘)

SubEqn

LHS-79^(∘)=RHS-79^(∘)

-15^(∘) = - x^(∘)

MultEqn

LHS * (-1)=RHS* (-1)

15^(∘) = x^(∘)

RearrangeEqn

Rearrange equation

x^(∘) = 15^(∘)