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Core Connections Algebra 2, 2013 View details

1. Section 12.1

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Exercise 7 Page 622

Practice makes perfect

a We are given the length of one leg and the measure of an acute angle of a right triangle. We want to find the length of the other leg.

Note that the given side is opposite the given angle, and the side we want to find is adjacent to the given angle. Therefore, we will use the cotangent ratio. cot θ = adjacent/opposite In our triangle, we have that θ = 24^(∘) and the length of the opposite leg is 36. We want to find the length of the leg adjacent to the angle.

tan θ = opposite/adjacent

SubstituteValues

Substitute values

tan 24^(∘) = x/36

▼

Solve for x

MultEqn

LHS * 36=RHS* 36

36 tan 24^(∘) = x

RearrangeEqn

Rearrange equation

x = 36 tan 24 ^(∘)

UseCalc

Use a calculator

x = 80.85732...

RoundDec

Round to 2 decimal place(s)

x≈ 80.86

Therefore, x is approximately 80.86 units.

b We are given the length of two sides and two interior angles of a triangle.

Notice that the given sides are opposite the given angle. Therefore, we will use the Law of Sines to find the value of x.

Law of Sines
For any triangle, the ratio of the sine of an angle to its opposite side is constant.

This law allows us to write the following equation. sin 23^(∘)/10 = sin 70 ^(∘)/x Let's solve this equation for x.

sin 23^(∘)/10 = sin 70 ^(∘)/x

▼

Solve for x

MultEqn

LHS * x=RHS* x

sin 23^(∘)/10 x = sin 70 ^(∘)

DivEqn

.LHS /sin 23^(∘)/10.=.RHS /sin 23^(∘)/10.

x = sin 70 ^(∘)/. sin 23^(∘) /10.

DivByFracD

a/b/c= a * c/b

x = 10 sin 70 ^(∘)/sin 23^(∘)

UseCalc

Use a calculator

x = 24.04959...

RoundDec

Round to 2 decimal place(s)

x ≈ 24.05

Therefore, x is approximately 24.05 units.

c This time we are given all side lengths of a triangle and we want to find the value of one the interior angles.

To find the value of x, we can use the Law of Cosines which relates the cosine of each angle to the side lengths of the triangle.

Let's relate the given lengths to x using this theorem. 3^2= 10^2+ 11^2-2( 10)( 11)cos x Finally, let's find the value of x by solving the above equation

3^2=10^2+11^2-2(10)(11)cos x

▼

Solve for x

CalcPowProd

Calculate power and product

9=100+121-220 cos x

AddTerms

Add terms

9=221-220 cos x

SubEqn

LHS-221=RHS-221

- 212 = - 220 cos x