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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

6. Secants, Tangents, and Angle Measures

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Exercise 50 Page 767

Find the length of each side to show that the triangle is a right triangle.

m∠ C ≈ 54.5^(∘)

Practice makes perfect

Let's start by drawing △ BCD, whose vertices are the given ones.

Before finding m ∠ C, let's find the length of each side by using the Distance Formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)First, we will find CD.

CD = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)

SubstitutePoints

Substitute ( -6,-5) & ( -1,2)

CD = sqrt(( -1-( -6))^2 + ( 2-( -5))^2)

▼

Simplify right-hand side

- (- a)=a

CD = sqrt((-1+6)^2 + (2+5)^2)

Add terms

CD = sqrt(5^2 +7^2)

Calculate power

CD = sqrt(25 + 49)

Add terms

CD = sqrt(74)

The computations to find the other two side lengths are summarized in the following table.

Points	d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)	Length
B( -1, -5) and C( -6, -5)	BC = sqrt(( -6-( -1))^2 + ( -5-( -5))^2)	BC = 5
B( -1, -5) and D( -1, 2)	BD = sqrt(( -1-( -1))^2 + ( 2-( -5))^2)	BD = 7

Since BC^2+BD^2 = CD^2, we conclude that △ BCD is a right triangle with ∠ B being the right angle.

Next, applying the tangent ratio, we can write the following equation involving the measure of ∠ C. tan C = BD/BC ⇒ tan C = 7/5 Finally, applying the inverse tangent and a calculator, we will find the required measure. m∠ C = arctan (7/5) ≈ 54.5^(∘)

Subchapter links

Exercises p.763-767