Exercise 53 Page 596

Practice makes perfect

a We are asked to draw an acute, scalene triangle

A B C

including an altitude of length

h

originating at vertex

A .

Let's do this. Remember that an altitude in a triangle is perpendicular to the base.

b In this part we are asked to use trigonometry to express

h

in terms of

m ∠ B .

To do this, let's recall that in a right triangle the sine of an angle is the ratio of a side length opposite this angle to the length of the hypotenuse. Having this definition in mind, let's look at our triangle.

Since an altitude divided

△ A B C

into two right triangles, we can write an equation for

sin B .

The leg opposite to this angle has a length of

h

and the hypotenuse is

\overline{A B} .

sin B = \frac{h}{A B}

Next, we will isolate

h

in the above equation.

sin B = \frac{h}{A B}

MultEqn

$LHS \cdot A B = RHS \cdot A B$

A B sin B = h

RearrangeEqn

Rearrange equation

h = A B sin B

c Now we will write an equation to find the area of a triangle

A B C .

Let's recall the formula for the area of a triangle.

A = \frac{1}{2} b h

In this formula,

b

is the length of the base and

h

is the length of an altitude of a triangle. In

△ A B C,

the altitude has a length of

h,

and the length of the base is

B C .

Therefore, we can write an equation for the area of

△ A B C .

A = \frac{1}{2} (B C) h

Since we are asked to use trigonometry in our equation, we will use the fact that we found that

h = A B sin B .

A = \frac{1}{2} (B C) h

Substitute

$h = A B s i n B$

A = \frac{1}{2} (B C) A B s i n B

d In this part we are given that

m ∠ B

4 7^{\circ},

A B = 11.1,

B C = 14.1,

and

C A = 10.4

and asked to evaluate the area of

△ A B C .

To do this, we can use the formula we found in Part C.

A = \frac{1}{2} (B C) A B sin B

SubstituteValues

Substitute values

A = \frac{1}{2} (14.1) (11.1) sin 4 7^{\circ}

Multiply

A = 78.255 sin 4 7^{\circ}

UseCalc

Use a calculator

A = 57.232 \dots

RoundDec

Round to $1$ decimal place(s)

A \approx 57.2

e To write an equation for the area of

△ A B C

using trigonometry in terms of a different angle measure, let's look at the picture for the last time.

First, we can express

h

in terms of

m ∠ C .

Again, we will use the definition of sine.

sin C = \frac{h}{A C} ⇓ h = A C sin C

Next, we will substitute the value of

h

into the area formula.

A = \frac{1}{2} (B C) h

Substitute

$h = A C s i n C$

A = \frac{1}{2} (B C) A C s i n C

Notice that this is only a sample solution.

Subchapter links

Check Your Understanding p.592

Practice and Problem Solving p.593-596

H.O.T. Problems p.596

Standardized Test Practice p.597

Spiral Review p.597

Skills Review p.597

Exercise 53 Page 596

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