## 17Calculus Precalculus - Similar Triangles

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Similar triangles show up a lot in calculus, especially when solving related rates and optimization problems in calculus.

For this discussion, we will use the two triangles below, Figures 1 and 2. We are told they are similar triangles. Since they are similar, we can rotate one until it looks like a smaller/larger version of the other one. They are already in the correct orientation here.

Since they are similar, we can move the smaller triangle into the larger one, until they are combined. This is shown in Figure 3, below. We do not have to show them like this but we think it makes it easier to see the equations we need to write.
Notice that we have combined the triangles by matching the top point at angle $$\gamma$$. However, we could have also combined them by overlapping either of the other two angles, $$\alpha$$ or $$\beta$$.

Figure 1 - large triangle

Figure 2 - small triangle

Figure 3 - combined triangles at angle $$\gamma$$

Notice in Figure 3, we used Greek letters to label the angles, $$\alpha$$ (alpha), $$\beta$$ (beta) and $$\gamma$$ (gamma). This is very common in calculus. For an explanation of the use of the Greek alphabet in calculus, see the 17calculus notation page.

There are two main ways to write the equations. They are equivalent, so you can write them the way that makes sense to you (check with your instructor to see what they expect).

Vertically

The first way is to write the similar sides vertically, in the same fraction.
$$\displaystyle{ \frac{A}{a} = \frac{B}{b} = \frac{C}{c} }$$
This way of writing the equation is shorthand for the three separate equations

 $$\displaystyle{ \frac{A}{a} = \frac{B}{b} }$$ $$\displaystyle{ \frac{A}{a} = \frac{C}{c} }$$ $$\displaystyle{ \frac{B}{b} = \frac{C}{c} }$$

Also, we could have written the reciprocal of each fraction, as long as the relationship stays the same, i.e. either all the large triangle lengths are in the numerator or all the small triangle lengths all have to be in the numerator.

Horizontally

The other way to write the ratios is to write the matching lengths across the equal sign (horizontally).

 $$\displaystyle{ \frac{A}{B} = \frac{a}{b} }$$ $$\displaystyle{ \frac{A}{C} = \frac{a}{c} }$$ $$\displaystyle{ \frac{B}{C} = \frac{b}{c} }$$

By cross multiplying, you see that the equations are the same. For example, looking at the first set of fractions, we have $$\displaystyle{\frac{A}{B} = \frac{a}{b} }$$. We can cross multiply the $$B$$ and $$a$$ to get $$\displaystyle{\frac{A}{a} = \frac{B}{b} }$$. This is the same fraction as in the vertically aligned case. The other two equations work similarly.

Practice

Calculate $$x$$ using these similar triangles.

Problem Statement

Calculate $$x$$ using these similar triangles.

$$x = 9$$

Problem Statement

Calculate $$x$$ using these similar triangles.

Solution

### 2874 video

$$x = 9$$

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Calculate $$x$$ and $$y$$ using these similar triangles.

Problem Statement

Calculate $$x$$ and $$y$$ using these similar triangles.

$$x = 45/2$$ and $$y = 32$$

Problem Statement

Calculate $$x$$ and $$y$$ using these similar triangles.

Solution

### 2875 video

$$x = 45/2$$ and $$y = 32$$

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Calculate $$x$$ using these similar triangles.

Problem Statement

Calculate $$x$$ using these similar triangles.

$$x = 15$$

Problem Statement

Calculate $$x$$ using these similar triangles.

Solution

### 2876 video

$$x = 15$$

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Really UNDERSTAND Precalculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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