Similarity, Right Triangles, & Trigonometry - Geometry

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Question

Complete the statement by choosing from one of the following

Regarding the triangle below, to find the area using the formula $A = \frac{1}{2}ab(sin(C))$ , I assume that

Answer

Recall that $sin(\theta) = \frac{opposite}{hypotenuse}$ for right triangles. When we draw a line down from the vertex to the opposite side this creates two right triangles within the original obtuse triangle. If we are solving for , then becomes our hypotenuse and becomes the opposite side we are working with. Therefore $sin(C) = \frac{h}{b}$ .

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Question

Given the triangle below, what is the formula for finding the area?

Answer

The formula for the area of a triangle is $\frac{1}{2}$ (base)(height). Since this is an obtuse triangle we need to break it into two right triangles by drawing the line down from the vertex perpendicular to the opposite side.

Now that we have two right triangles we can solve for the area of this triangle. Notice our base is and our height is . Plugging into the formula for the area of a triangle gives us $A = \frac{1}{2}ah$ . Most of the time, we will not have an exact length for , but we may have, or will be able to solve for, the lengths of and the angles . Using our relationship for right triangles, we know $sin(\theta) = \frac{opposite}{hypotenuse}$ . In this case we will use as our angle. So

$sin(C) = \frac{h}{b}$

We can plug in for in our formula for area and we are left with $A = \frac{1}{2}ab(sin(C))$ .

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Question

Given the triangle below, find the area and round to the second decimal place (in degrees).

Answer

Following from the figure above, we draw an auxiliary line from the vertex perpendicular to the opposite side. We will call this line for height and will label the new vertex . The formula we will use is $A = \frac{1}{2}ab(sin(C))$ . First we begin by labeling our variables. We will let be the base of our triangle and be the hypotenuse of the triangle formed by . The values for the variables are as follows:

Plugging these into our formula, it follows that:

$A = \frac{1}{2}(8)(6)sin(50)$

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Question

Using the triangle below, find the area then solve for .

Answer

To solve for , we first need to find the area using our formula $A = \frac{1}{2}ab(sin(C))$ , considering the triangle . We will begin by labeling our variables:

To find the area, we plug the values for the variables into the formula.

$A = \frac{1}{2}(12)(8)sin(30)$

Knowing that our area is 24, we can now solve for . Recall that:

$sin(C) = \frac{h}{b}$

$\Rightarrow bsin(c) = h$

It follows that:

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

$A = \frac{1}{2}ah$ (because )

$24 = \frac{1}{2}(12)h$ (plugging in our values for area and )

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Question

Is the following statement True or False?

We want to use the formula $A = \frac{1}{2}ab(sin(C))$ . Consider an obtuse triangle . We know the lengths of and , but only know the angle for . We are still able to use this formula.

Answer

There are two approaches to this problem. We are able to calculate the angle by using the Sine Law. The Sine Law states:

$\frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)}$

So we can set $\frac{b}{sin(B)} = \frac{c}{sin(C)}$ and solve accordingly for angle .

Our other option is to use the area formula we have been but altering it to correspond to angle . We would draw our vertical line down from the vertex as shown below and our formula would be in the form $A = \frac{1}{2}ac(sin(B))$ .

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Question

Solve for x using the formula $A = \frac{1}{2}ab(sin(C))$ given that the area of the following triangle is (round to the second decimal place if needed).

Answer

Even though the formula $A = \frac{1}{2}ab(sin(C))$ is using sides and angle , this is a general formula and can be used with any angle in the triangle. Since we are now working with an obtuse angle rather than an acute angle, we need to do some more work to get the logic right.

Using the figure above, to be able to label the sides we are using correctly, we extend the original triangle horizontally past the obtuse angle and draw a vertical line down from the top vertex to form a right angle. This vertical line is . Angle for the supplementary (orange) triangle is . Using the fact that $sin(A) = \frac{h}{b} = \frac{h}{2x}$ , we can set up our formula to be the following:

(either angle can be used and I will show this to be true)

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

$20 = \frac{1}{2}(x)(2x)sin(60)$ (Using from the original triangle)

$\frac{20}{sin(60)} = x^2$

This shows that when using this formula for an obtuse angle, you can use either the supplementary angle you made, or the original. It is always helpful to draw this supplementary triangle in order to be able to visualize and understand logically how the formula is working for obtuse angles as well.

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Question

Stephanie is building a triangular garden in her backyard. She has cut wood for the sides with lengths of 5, 4, and 3 feet. Knowing that the height from the vertex $A$ to the base formed by the line $CB$ is 1.778 feet, first find the area that will be in the garden and then calculate each angle that Stephanie will have to put the boards together in order for the lengths to fit just right. Round your answer to the third decimal place if needed.

Answer

We must begin by finding the area first. To find the area we begin by using our equation $A = \frac{1}{2}ab(sin(C))$ and recall that $sin(C) = \frac{h}{b}$ . We will let:

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

$A = \frac{1}{2}ab\frac{h}{b} \text{because $sin(C) = \frac{h}{b}$}$

$A = \frac{1}{2}ah$

$A = \frac{1}{2}(6)(1.778)$

So our area is

How to find $\angle C$ : , ,

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

$\frac{2A}{ab} = sin(C)$

$sin^{-1}(\frac{2A}{ab}) = C$

$sin^{-1}(\frac{2(5.334)}{(6)(4)} = C$

$sin^{-1}(\frac{10.668}{24} = C$

How to find $\angle B$ : , ,

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

$\frac{2A}{ab} = sin(C)$

$sin^{-1}(\frac{2A}{ab}) = C$

$sin^{-1}(\frac{2(5.334)}{(6)(3)}) = C$

To find $\angle A$ , we know the angles of a triangle should add up to 180 so:

$180 = \angle A + \angle B + \angle C$

$180 = 26.391 + 36.346 + \angle C$

$117.263 = \angle C$

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Question

Is the following statement True or False?

In order to use this formula, you only need to know two of the triangles sides' lengths.

Answer

In order to use this formula you need to either the height (which can be used to find the angle) or the angle (which can be used to find the height).

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Question

Using the information from the following triangle, find the area. Round to the second decimal place.

Answer

Using the formula $A = \frac{1}{2}ab(sin(C))$ , we will let and

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

$A = \frac{1}{2}(8)(10)sin(130)$

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Question

You are making your friend a triangular fabric decoration for their home. It has ripped so you need to go buy new fabric. The fabric you are buying is \$3 per square foot. Using the dimensions below, how much money do you need to buy enough fabric?

Answer

Using the formula $A = \frac{1}{2}ab(sin(C))$ , we will let and .

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

$A = \frac{1}{2}ab\frac{h}{b}$ (because $sin(C) = \frac{h}{b}$ )

$A = \frac{1}{2}(18)(10)\frac{6}{10}$

Then we must account for each square foot costing \$3.

The answer is \$162.

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Question

The Law of Cosines is a generalization of which common theorem?

Answer

The Law of Cosines is a generalization of the Pythagorean Theorem. To better understand this, start by drawing triangle ABC with sides a, b, and c. Then, construct the altitude from point A to get height h. Finally label the distance between point B and the point where h meets side a as distance r.

We will arbitrarily work to solve for side b, but we could do the same for a and c (after drawing the correct altitudes), which is why we can generalize that the other two formulas to find a and c follow.

By definition, we know that $\sin\angle B= \frac{h}{c}$ . If we multiply both sides by c, we get $c\sin\angle B= h$ .

Also, we know that $\cos\angle B= \frac{r}{c}$ . If we multiply both sides by c, we get $c\cos\angle B= r$ .

Using the Pythagorean Theorem, we get .

Substituting in h and r from above, we get:

$b^2=(c\sin \angle B)^2+ (a-c\cos\angle B)^2$

$= c^2\sin^2 \angle B + a^2-2ac\cos\angle B + c^2\cos^2\angle B$

$= c^2 (sin^2 \angle B+cos^2\angle B) + a^2-2ac\cos\angle B$ (because $sin^2 \angle B+cos^2\angle B=1$ )

$= c^2 + a^2-2ac\cos\angle B$

Therefore $b^2= c^2 + a^2-2ac\cos\angle B$ .

The other two versions of Law of Cosines are:

$a^2= b^2 + c^2-2bc\cos\angle A$

$c^2= a^2 + b^2-2ab\cos\angle C$

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Question

Choose the answer choice that correctly states the Law of Sines, given the triangle below.

Answer

The Law of Sines uses ratios of a triangle’s sides and angles to allow us to be able to solve for unknown sides and/or angles. This relationship is true for any triangle.

The Law of Sines is $\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$ .

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Question

Which of the following expressions correctly states the Law of Cosines? Use the following triangle for reference.

Answer

The Law of Cosines should look similar to a theorem we use quite commonly for right triangles. The Law of Cosines is basically the Pythagorean Theorem, but it has an added argument so that it is true for any and all triangles.

The Law of Cosines is $A^{2}+B^{2}-2AB\cos (\gamma )=C^{2}$

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Question

Choose the answer choice that correctly states the Law of Sines, given the triangle below.

Answer

The Law of Sines uses ratios of a triangle’s sides and angles to allow us to be able to solve for unknown sides and/or angles. This relationship is true for any triangle.

The Law of Sines is $\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$ .

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Question

True or false: The Law of Sines applies to all types of triangles.

Answer

This is true. While the Pythagorean Theorem only applies to right triangles, the Law of Sines applies to all triangles no matter if they are right, acute, or obtuse.

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Question

True or False: The Law of Cosines applies to all triangles.

Answer

This is true. While the Pythagorean Theorem only applies to right triangles, the Law of Cosines applies to all triangles no matter if they are right, acute, or obtuse.

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Question

Assume the following two triangles are similar. Using the fact that their side ratios are $\frac{AB}{DE}= \frac{AC}{DF}$ , what trigonomic function could this represent for angles and ?

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{AB}{DE}= \frac{AC}{DF}$

$\begin{tabular}{lcc} \ $\implies AB = \frac{DE*AC}{DF}$ & $multiply ; both ; sides ; by; DE$ \ $\implies \frac{AB}{AC} = \frac{DE}{DF}$ & $divide ; both ; sides ; by ; AC$ \ \end{tabular}$

If we look at angle , we see that $\frac{AB}{AC}$ is the $\frac{opposite}{hypotenuse}$ . Looking at angle we see that $\frac{DE}{DF}$ is also the $\frac{opposite}{hypotenuse}$ . This is the definition of the sine function. So:

$sin(C) = \frac{AB}{AC} = \frac{DE}{DF} = sin(F)$

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Question

Using the information of the side lengths in the triangle below, use the side ratios to find the .

Answer

The definition of cosine of an acute angle is $cos(angle) = \frac{adjacent}{hypotenuse}$ . So here we must determine what sides are adjacent and which is the hypotenuse. The hypotenuse is the easiest to pick out. This is the side that is directly across from the right angle in a right triangle. Our hypotenuse is . Now we must choose the adjacent side. Adjacent means next to, and since $\bar{AC}$ is our hypotenuse then $\bar{AB}$ must be our adjacent side. So:

$cos(A) = \frac{adjacent}{hypotenuse}$

$\implies cos(A) = \frac{AB}{AC}$

$\implies cos(A) = \frac{3}{4}$

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Question

Assume the two triangles below are similar. Using the fact that their sides ratios are $\frac{BC}{EF} = \frac{AC}{DF}$ , what trigonometric function could this represent for angles and

Answer

We must begin by manipulating our equation of the side ratios to get fractions that include both of the sides of the same triangles:

$\frac{BC}{EF} = \frac{AC}{DF}$

$\implies BC = \frac{EF*AC}{DF}$ multiply both sides by

$\implies \frac{BC}{AC} = \frac{EF}{DF}$ divide both sides by

If we look at angle we see that $\frac{BC}{AC}$ is the $\frac{adjacent}{hypotenuse}$ . Looking at angle we see that is also $\frac{adjacent}{hypotenuse}$ . This is the definition of the cosine function of an angle. So:

$cos(C) = \frac{BC}{AC} = \frac{EF}{DF} = cos(F)$

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Question

Using the sides of a right triangle, what is the definition of $tan(\theta)$ ?

Answer

We can see this to be true by looking at the following triangle.

We are considering $\angle C$ . The opposite side to $\angle C$ is $\bar{AB}$ . $\angle C$ has two adjacent sides, but since $\bar{AC}$ is opposite to the right angle, this is the hypotenuse of the triangle. So the adjacent side to $\angle C$ must be $\bar{BC}$ . We can set up the following equation to check and make sure that $tan(\theta)=\frac{opposite}{adjacent}$ .

$tan(\theta)=\frac{opposite}{adjacent}$

$tan(36.87) = \frac{3}{4}$

This shows that $tan(\theta)=\frac{opposite}{adjacent}$ .

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