Understanding Law of Cosine: Solving Triangles with Sines and Cosines

$m ∠ C$	$c^{2} = a^{2} + b^{2} - 2 a b cos C$	Relationship between $a^{2} + b^{2}$ and $c^{2}$	Conclusion
$m ∠ C < 9 0^{\circ}$	$c^{2} = a^{2} + b^{2} - 2 a b > 0 cos C$	$c^{2} < a^{2} + b^{2}$	If $∠ C$ is acute, not too many conclusions can be made. The opposite side to $∠ C$ can be the largest side, the shortest side, or none.
$m ∠ C = 9 0^{\circ}$	$c^{2} = a^{2} + b^{2} - 2 a b = 0 cos 9 0^{\circ}$	$c^{2} = a^{2} + b^{2}$	The cosine of $9 0^{\circ}$ is $0 .$ In this case, the Law of Cosines becomes the Pythagorean Theorem. This means that the opposite side to $∠ C$ is the largest side of the triangle.
$9 0^{\circ} < m ∠ C < 18 0^{\circ}$	$c^{2} = a^{2} + b^{2} - 2 a b < 0 cos C$	$c^{2} > a^{2} + b^{2}$	If $∠ C$ is obtuse, its measure is greater than the measures of $∠ A$ and $∠ C .$ Therefore, its opposite side is the largest side of the triangle.

m ∠ C

c^{2} = a^{2} + b^{2} - 2 a b cos C

Relationship between

a^{2} + b^{2}

and

c^{2}

Conclusion

m ∠ C < 9 0^{\circ}

c^{2} = a^{2} + b^{2} - 2 a b > 0 cos C

c^{2} < a^{2} + b^{2}

∠ C

is acute, not too many conclusions can be made. The opposite side to

∠ C

can be the largest side, the shortest side, or none.

m ∠ C = 9 0^{\circ}

c^{2} = a^{2} + b^{2} - 2 a b = 0 cos 9 0^{\circ}

c^{2} = a^{2} + b^{2}

The cosine of

9 0^{\circ}

0 .

In this case, the Law of Cosines becomes the Pythagorean Theorem. This means that the opposite side to

∠ C

is the largest side of the triangle.

9 0^{\circ} < m ∠ C < 18 0^{\circ}

c^{2} = a^{2} + b^{2} - 2 a b < 0 cos C

c^{2} > a^{2} + b^{2}

∠ C

is obtuse, its measure is greater than the measures of

∠ A

and

∠ C .

Therefore, its opposite side is the largest side of the triangle.

	$Speed = \frac{Distance}{Time}$
Length	$S D$	$D N$
Substitution	$330 = \frac{S D}{3}$	$330 = \frac{D N}{2}$
Calculation	$S D = 990 mi$	$D N = 660 mi$

Speed = \frac{Distance}{Time}

Length

S D

D N

Substitution

330 = \frac{S D}{3}

330 = \frac{D N}{2}

Calculation

S D = 990 mi

D N = 660 mi

Kevin wants to combine his favorite sports of ice skating and windsurfing into one activity. He calls it ice surfing.

Kevin has designed a new sail, specifically for windsurfing, with the following dimensions.

two quadrilaterals that are mirror images of each other in one side

Before he buys the fabric to make the sail, he needs to know its area. What is the area of the sail? The fabric store only sells material by the meter and rounded to one decimal place.

Examining the sail, consider what may help us find the total area. Well, treating the support beam as a line of reflection, we can see that the parts above and below are mirror images of each other.

That means if we can determine the area of one part of the sail, then we can double its area to find the total area of the sail. For now, let's work with the bottom half. We will add a diagonal to that part to split it into two triangles and find their areas.

Since one of the triangles is a right triangle and we know both of its leg measurements, we can find its area by using the formula for the area of a triangle. A=1/2( 3.30)( 0.85)= 1.4025 m^2 Next, we need to find the area of the second triangle. Again, because we know both of its leg measurements, we can find the length of the side labeled c by using the Pythagorean Theorem.

Now we can find either of its three angles by using the Law of Cosine.

Now let's apply the formula for calculating a triangle's area using sine to find its area.

Finally, we have the information we need to determine the total area of the sail. First, let's add the area of the two triangles. Recall that the two parts of the whole sail are a mirror image of each other. Therefore, we can multiply the sum of the two triangles by 2 to account for the entire sail. 2(1.4025+2.34340...)≈ 7.5m^2

What is the perimeter of the triangle in centimeters? Round the answer to one decimal place.

a triangle with three sides of 4.1 cm, 2.2 cm, and x cm

To determine the perimeter, we need to calculate the length of all sides. Currently, we do not know the length of x. Since we know two side lengths and an angle, we can use an equation from the Law of Cosines to determine x. Notice that x is not on the opposite side of the known angle A. Therefore, x goes on the right-hand side of the formula.

To solve this we will use the Quadratic Formula.

We cannot continue this simplification without a graphing calculator. Let's calculate 4.4cos 72^(∘).

Let's replace 4.4cos 72^(∘) by 1.35967... and continue.

When we know x, we can find the perimeter by adding all of the side lengths. P= 4.20576...+4.1+2.2≈ 10.5 cm

What is the perimeter of the quadrilateral $A B C D$ in inches? Give the answer in exact form.

polygon ABCD which consists of an equilateral triangle ADB and another triangle BDC which sits on one of the sides

To determine the perimeter of ABCD we must know all of its sides. Notice that the figure consists of two triangles, △ ABD and △ BDC. In △ BDC we know two sides and the included angle. This is enough information to calculate the length of BD by using the Law of Cosine.

Notice that △ ABD is an equilateral triangle. Therefore, it has three congruent sides. Since BD is sqrt(4.75), we know that the remaining sides have the same length.

Finally, we will add AD, DC, CB, and AB to determine the perimeter of ABCD. Notice that BD is not part of the perimeter, as it is inside the shape. We will keep the answer in exact form. sqrt(4.75)+ 2.5+ 1+ sqrt(4.75) ⇓ 2sqrt(4.75)+3.5

Solving Triangles Using the Law of Cosines

Catch-Up and Review

Investigate Distances in Space Using Trigonometry

Limitations of the Law of Sines

Law of Cosines

Proof

Proof

Knowing Two Sides and the Included Angle of a Triangle

Hint

Solution

Knowing Three Sides of a Triangle

Hint

Solution

Finding $m ∠ L$

Finding $m ∠ M$

Finding $m ∠ K$

Investigating the Cosine Ratio of Obtuse Angles

Relationship Between the Angles of a Triangle

Solving Problems Using the Law of Cosines

Hint

Solution

Finding $S D$ and $D N$

Finding $S D$

Calculating Distances in Space Using Trigonometry

Hint

Answer

Solving Triangles Using the Law of Cosines

Recommended exercises

	10 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Solving Triangles Using the Law of Cosines

Catch-Up and Review

Proof

Proof

Hint

Solution

Hint

Solution

Finding m∠L

Finding m∠M

Finding m∠K

Hint

Solution

Finding SD and DN

Finding SD

Hint

Answer

Solving Triangles Using the Law of Cosines

Recommended exercises

Finding $m ∠ L$

Finding $m ∠ M$

Finding $m ∠ K$

Finding $S D$ and $D N$

Finding $S D$