Mathematics for Technology I (Math 1131) Durham College, Mathematics
Free • 55 lessons
• 1 quizzes
• 10 week duration
• Numerical Computation

Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.

• Measurements

An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

• Trigonometry with Right Triangles

Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.

• Trigonometry with Oblique Triangles

This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.

• Geometry

This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.

Mathematics for Technology I (Math 1131)

Solve problems involving circles

Let’s review the anatomy of a circle: • Radius (r) is the distance from the center of the circle to the closed curve.
• Diameter (d) is the distance measured across the circle through the center. The diameter is often represented at d = 2⋅r since it’s twice the radius.
• Central angle is one whose vertex is at the center of the circle.
• Circumference (C) is the distance (perimeter) around the circle.

Circumference

To find the circumference of a circle, the following two formulas can be used:

By rearranging the formulas, the definition of π (Greek letter pi) can be derived as the ratio of the circumference of any circle to its diameter:

Rearranging further, we can also find the diameter:

Area

For a circle of radius r, the following two formulas can be used to the area:

Question:   The area of a circle is 583 cm². Find the radius.

You can use either of the formulas, but we’ll use formula 1.

$A=\pi {r}^{2}\phantom{\rule{0ex}{0ex}}583={\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}\frac{583}{\mathrm{\pi }}={\mathrm{r}}^{2}\phantom{\rule{0ex}{0ex}}\sqrt{\frac{583}{\mathrm{\pi }}}=\mathrm{r}\phantom{\rule{0ex}{0ex}}±13.6=\mathrm{r}\phantom{\rule{0ex}{0ex}}$

Since radius cannot be negative, only +13.6 cm is correct to the correct number of significant figures.

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Arc and Sector • An inscribed angle is one whose vertex is on the circle.
• An arc is a portion of the circle between two points on the circle.
• A sector is a plane region bounded by two radii and one of the arcs intercepted by those radii (pizza slice).
• A Chord is a line segment spanning the closed curve (shown below). Properties of Intersecting Chords If two chords in a circle intersect, the product of the parts of one chord is equal to the product of the parts of the other chord. For example, given the circle and two chords shown below, we can find x by setting up the following equation:

$16.3·x=12.5\left(10.1\right)\phantom{\rule{0ex}{0ex}}$

Solving for x, we get:

$\frac{12.5\left(10.1\right)}{16.3}=x⇒7.75\phantom{\rule{0ex}{0ex}}$

Properties of Tangents and Secants

A secant is a straight line that cuts the circle in two points and produces a chord lying within the circle. A tangent is a straight line that touches a circle at just one point. There are two rules about tangents and circles that you need to be aware of:

1. A tangent is perpendicular (at 90°) to the radius drawn through the tangent point.
2. Two tangents drawn to a circle from a point outside the circle are equal, and they make equal angles with a line from the point to the center of the circle (shown below). Properties of Semicircles

Any angle inscribed in a semicircle is a right angle.

Question:   Find the distance x in the figure below. Solution:   According to the rule, angle θ has to be 90°. Therefore, this generates a right triangle to which you can use Pythagorean theorem to solve for the missing sides. The blue arrow tells use the radius is 125 units long. Given that this is a semi-circle, we can multiply 125 by 2 to find a second missing side – the hypotenuse.

Let’s look at a few more challenging examples where we use the Pythagorean theorem along with the properties of circles to solve problems: