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The area of a triangle is half the product of its base and the corresponding height. In this lesson, a formula for the area of a triangle involving the sine ratio will be derived.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Izabella has a triangular table in her kitchen. She wants to cover the surface of the table with a tablecloth. Two side lengths and two angle measures of the table are given.

How much fabric does Izabella need to cover the surface of the table?Typically to find the area of a triangle, the measurements of its height and base are used. On the other hand, if the lengths of two sides and the measure of their included angle are known, what should be the process to find the area of the triangle?

In $△ABC,$ the lengths of $AB$ and $BC$ are $6$ and $8$ inches, respectively. The measure of $∠ABC$ varies from $15_{∘}$ to $90_{∘}.$ What information is needed to find the area of $△ABC?$ How can the measure of $∠ABC$ be used to find the area of $△ABC?$

One way of finding the area of an equilateral triangle is by using trigonometric ratios.

Izabella wants to finish her math homework before she goes out to buy some fabric for her table. She is given an equilateral triangle whose side length is $1$ unit.

Izabella's homework problem is to find the area of this triangle. Help her find the answer.

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What information is needed to find the area of a triangle?

The area of a triangle is half the product of its base and the corresponding height. Therefore, the height of $△ABC$ needs to be found. Suppose line $ℓ$ bisects $∠A$ and intersects $AB$ at point $M.$

Since $△ABC$ is an equilateral triangle, the measure of each of its interior angles is $60_{∘}.$ Consequently, $∠ACM$ measures $30_{∘}.$

Considering $△ACM,$ it can be concluded that $∠AMC$ is a right angle by the Triangle Angle Sum Theorem.

Consequently, $CM$ is an altitude of $△ABC$ and $AB$ its corresponding base. To find $CM,$ the right triangle whose vertices are $A,$ $M,$ and $C$ will be considered. Note that the hypotenuse of $△AMC$ is $1,$ and that $CM$ is the opposite side to the angle whose measure is $60_{∘}.$ Therefore, the sine ratio can be used to find $CM.$$sin60_{∘}=1CM ⇒sin60_{∘}=CM $

Recall that the exact value of $sin60_{∘}$ is $23 .$ $sin60_{∘}=CM⇒CM=23 $

Now that the height was found, the area of $△ABC$ can be calculated.
The area of the equilateral triangle is $43 units_{2}.$ In the previous example, it was shown that trigonometric ratios can be used to find the area of a triangle given the lengths of two sides and the measure of their included acute angle. What happens if the included angle is obtuse?

In $△ABC,$ the lengths of $AB$ and $BC$ are $6$ and $8$ inches, respectively. The measure of $∠ABC$ varies from $15_{∘}$ to $165_{∘}.$ How can the measure of $∠ABD$ be used to find the area of $△ABC$ when $∠ABC$ is obtuse? What is the relation between the areas of $△ABC$ when the measure of $∠ABC$ is $50$ degrees and $130$ degrees? What is the relation between the sine ratio of an acute angle and its supplementary angle?

Izabella finished her homework and is ready to go. On her way out, she notices an empty triangular region in the garden.

Her father wants to fill this region with topsoil. To figure out how much topsoil he needs, he asks Izabella to find the area of the region. Help them find the area. If it is necessary, round the answer to the nearest square foot.

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Begin by drawing the external altitude of the $40-foot$ side.

Begin by drawing the external altitude of the $40-foot$ side.

Notice that the angle opposite the altitude is the supplementary angle of $120_{∘}.$ Its measure can be found by subtracting $120_{∘}$ from $180_{∘}.$$180_{∘}−120_{∘}=60_{∘} $

The triangle formed by the altitude, the side whose length is $30$ feet, and the extension of the base, is a right triangle. Here, the hypotenuse is $30$ feet and the measure of the opposite angle to the altitude is $60_{∘}.$ Therefore, to find the height $h,$ the sine ratio can be used.
$sin60_{∘}=30h $

The above equation can be solved for $h.$
$sin60_{∘}=30h $

Solve for $h$

MultEqn

$LHS⋅30=RHS⋅30$

$30⋅sin60_{∘}=h$

RearrangeEqn

Rearrange equation

$h=30⋅sin60_{∘}$

UseCalc

Use a calculator

$h=2.598076…$

$A=21 bh$

SubstituteII

$b=40$, $h=2.598076…$

$A=21 (40)(2.598076…)$

Evaluate right-hand side

MoveRightFacToNumOne

$b1 ⋅a=ba $

$A=240 (2.598076…)$

CalcQuot

Calculate quotient

$A=20(2.598076…)$

UseCalc

Use a calculator

$A=51.961524…$

RoundInt

Round to nearest integer

$A≈52$

There is another way of finding the area of a triangle that can be derived from the previous examples.

The area of a triangle is half the product of the lengths of any two sides and the sine of their included angle. There are three possible formulas for every triangle.
### Proof

The first formula was obtained. To obtain the second formula, notice that $△ABD$ is also a right triangle. Therefore, the sine ratio can be applied again, this time to connect $h$ and $∠A.$

The proof of this theorem will be developed using the triangle shown, where $∠B$ is obtuse. However, the same proof is valid for all triangles.

To find the first formula, start by drawing the altitude from $B$ and let $h$ be its length. Since the altitude is perpendicular to the base, it generates two right triangles.

Because $△BCD$ is a right triangle, the height of the triangle can be related to the sine of $∠C$ using the sine ratio.$sinC=ah ⇔h=asinC $

Next, substitute the expression found for $h$ into the general formula for the area of a triangle. $Area=21 bh⇓Area=21 absinC $

$sinA=ch ⇔h=csinA $

By substituting this expression into the general formula for the area of a triangle, the second formula can be obtained. $Area=21 bh⇓Area=21 bcsinA $

To deduce the third formula, the altitude from $C$ or $A$ should be drawn. In this case, the altitude from $C$ will be arbitrarily drawn and labeled $D$ with a length of $h.$ Because $∠B$ is obtuse, the altitude will lie *outside* the triangle.

$sinCBD=ah $

Since $∠CBD$ and $∠B$ form a linear pair, they are supplementary angles. Recall that the sine of an angle is equal to the sine of its supplementary angle. With this information, and using the Substitution Property of Equality, a formula connecting $∠B$ and $h$ can be written. $⎩⎪⎪⎨⎪⎪⎧ sinB=sinCBDsinCBD=ah ⇒sinB=ah $

Multiplying both sides of the last equation by $a,$ it is obtained that $h=asinB.$ Finally, substitute this expression for $h$ into the formula for the area of $△ABC.$ $Area=21 ch⇓Area=21 acsinB $

While sitting on the bus on the way to the fabric store, Izabella began daydreaming about a formula for the area of a triangle involving the sine ratio. She dreams about applying this formula to find the area of the Bermuda Triangle which she remembers from the movie *Gulliver's Travels*.

Given the lengths of two sides and the measure of their included angle, calculate the area of the Bermuda Triangle. Round the answer to the nearest square mile.

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Use the formula for the area of a triangle involving the sine ratio.

Recall the formula for the area of a triangle involving the sine ratio.
Therefore, the area of the Bermuda Triangle is approximately $509456$ square miles.

$Area=21 absinC $

Since the lengths of two sides and the measure of their included angle are known, $a=1110,$ $b=980,$ and $C=69.5$ can be substituted into the formula to find the area.
$Area=21 absinC$

SubstituteValues

Substitute values

$Area=21 (1110)(980)sin69.5$

Evaluate right-hand side

Multiply

Multiply

$Area=21 (1087800)sin69.5$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$Area=21087800 sin69.5$

CalcQuot

Calculate quotient

$Area=543900sin69.5$

UseCalc

Use a calculator

$Area=509456.003732…$

RoundInt

Round to nearest integer

$Area≈509456$

In the following triangles, the lengths of two sides and the measure of their included angle is given. By using the formula derived earlier, find the area of the triangles. Round the answer to one decimal place.

By using a formula derived in this lesson, the challenge presented at the beginning can be solved. Recall that the question asked what amount of fabric Izabella needs to cover the triangular table.

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Begin by finding the included angle. Then use the formula for the area of a triangle using the sine ratio.

To find how much fabric Izabella needs, the surface area of the table must be found. Therefore, begin by finding the included angle.

To do so, the Triangle Angle Sum Theorem can be used.$x_{∘}+17_{∘}+54_{∘}=180_{∘}⇓x_{∘}=109_{∘} $

The lengths of two sides and the measure of their included angle are known. Therefore, the formula for the area of a triangle including the sine ratio can be used.
$A=21 (5)(4)sin109_{∘}$

Evaluate right-hand side

Multiply

Multiply

$A=21 (20)sin109_{∘}$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$A=220 sin109_{∘}$

CalcQuot

Calculate quotient

$A=10sin109_{∘}$

UseCalc

Use a calculator

$A=9.455185…$

RoundDec

Round to $1$ decimal place(s)

$A≈9.5$