McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
6. Secants, Tangents, and Angle Measures
Continue to next subchapter

Exercise 34 Page 766

Find mBD using the fact that AE and AC are secant to the circle and intersect each other outside it. Then, use the Arc Addition Postulate to rewrite mBC.

mDE=98^(∘)

Practice makes perfect

Let's begin by making a diagram where we label the given measures.

Since BC is a diameter of the circle we have that mBC=180^(∘). Additionally, the Arc Addition Postulate allows us to write the following equation. mBC = m BD + m DE + m EC To find m BD, we use the fact that AE and AC are secant to the circle and they intersect each other outside it. Thus, we can write the following equation. m∠ A = 1/2(m EC - m BD) Next, we substitute the corresponding measures into the equation above and solve it for m BD.
m∠ A = 1/2(m EC - m BD)
26^(∘) = 1/2( 67^(∘) - m BD)
Solve for m BD
52^(∘) = 67^(∘) - m BD
-15^(∘) = - m BD
15^(∘) = m BD
m BD = 15^(∘)
Finally, we substitute the three corresponding arc measures into the equation written at the beginning and solve it for m DE.
mBC = m BD + m DE + m EC
180^(∘) = 15^(∘) + m DE + 67^(∘)
Solve for m DE
180^(∘) = 82 + m DE
98^(∘) = m DE
m DE = 98^(∘)