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Core Connections Integrated II, 2015

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Core Connections Integrated II, 2015 View details

2. Section 6.2

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Exercise 56 Page 340

Notice that this is a right isosceles triangle.

x≈ 11.3

Practice makes perfect

Since this is a right triangle, we can calculate the length of its legs with the Pythagorean Theorem.

a^2+b^2=c^2

SubstituteValues

Substitute values

x^2+x^2=16^2

▼

Solve for x

Calculate power

x^2+x^2=256

Add terms

2x^2=256

.LHS /2.=.RHS /2.

x^2=128

sqrt(LHS)=sqrt(RHS)

x=± sqrt(128)

x > 0

x= sqrt(128)

Use a calculator

x= 11.31370...

Round to 1 decimal place(s)

x ≈ 11.3

In addition to being a right triangle it is also an isosceles triangle, which means it has two base angles of 45^(∘).

Now that we know the base angles, we can find the value of x by using the Law of Sines.

sin A/a=sin B/b

SubstituteValues

Substitute values

sin 90^(∘)/16=sin 45^(∘)/x

▼

Solve for x

LHS * x=RHS* x

sin 90^(∘)/16* x= sin 45^(∘)

LHS * 16=RHS* 16

sin 90^(∘)* x= 16 sin 45

.LHS /sin 90^(∘).=.RHS /sin 90^(∘).

x= 16 sin 45^(∘)/sin 90^(∘)

Use a calculator

x= 11.31370...

Round to 1 decimal place(s)

x≈ 11.3

Subchapter links

6.2.1 At Your Service p.340

6.2.2 Angles on a Pool Table p.343-344

6.2.3 Shortest Distance Problems p.347-348

6.2.4 The Number System and Deriving the Quadratic Formula p.353

6.2.5 The Number System and Deriving the Quadratic Formula p.355-356

6.2.6 Analyzing a Game p.359-360