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Showing posts with label Review. Show all posts
Showing posts with label Review. Show all posts

Tuesday, May 4, 2010

Review

Law of sines

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,


Law of cosines

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


Area of a triangle
Area =  \frac{1}{2}ab\sin \gamma = \frac{1}{2}bc\sin \alpha  = \frac{1}{2}ca\sin \beta



Area of a triangle using Heron's formula

(we did not learn this one in class but i thought it might be useful)

You can use herons formula if you are given all sides and no angles

Area = \sqrt{s(s-a)(s-b)(s-c)}

where s= \frac{a+b+c}{2} 

Review

Remember SOH CAH TOA to solve right triangles.


Also remember special right triangles.

Review of the Ambiguous Case


SAS - Side, Angle, Side
ASA - Angle, Side, Angle
AAS - Angle, Angle, Side
SSS - Side, Side, Side
HL - Hypotenuse Leg for Right Triangles.

We also discovered that SSA did not work to prove triangles congruent.
We politely called it the Donkey Theorem ; - )

By definition, the word ambiguous means open to two or more interpretations.
Such is the case for certain solutions when working with the Law of Sines.

• If you are given two angles and one side (ASA or AAS),
the Law of Sines will nicely provide you with ONE solution
for a missing side.
Unfortunately, the Law of Sines has a problem dealing with SSA.
If you are given two sides and one angle (where you
must find an angle), the Law of Sines could possibly provide you
with one or more solutions, or even no solution.

Before we investigate this situation, there are a few facts we need to remember.

Facts we need to remember:
1. In a triangle, the sum of the interior angles is 180º.
2. No triangles can have two obtuse angles.
3. The sine function has a range of .
4. If the = positive decimal can lie in the first quadrant (acute <) or in the second quadrant (obtuse <).


Let's look at some cases. In each example, decide whether the given information points to the existence of one triangle, two triangles or no triangles.

Example 1: In , a = 20, c = 16, and m = 30º. How many distinct triangles can be drawn given these measurements?

Use the Law of Sines:

C = sin-1 (0.4) = 24º (to the nearest degree) - in Quadrant I.
Sine is also positive in Quadrant II. If we use the reference angle 24º in Quadrant II,
the angle C is 156º.
But, with m
= 30º and m = 156º the sum of the angles would exceed 180º.
Not possible!!!!

Therefore, m = 24º, m = 30º, and m= 126º and only ONE triangle is possible.

Example 2: In , a = 7, c = 16, and m = 30º. How many distinct triangles can be drawn given these measurements?

Use the Law of Sines:

Since sin C must be 1, no angle exists for angle C.
NO triangle exists for these measurements.

Example 3: In , a = 10, b = 16, and m = 30º. How many distinct triangles can be drawn given these measurements?

Use the Law of Sines:

B = sin-1(.8) = 53.13010 = 53º.
Angles could be 30º, 53º, and 97º : sum 180º
The angle from Quadrant II could create angles 30º, 127º, and 23º : sum 180º



TWO triangles
possible.

This example is the Ambiguous Case. The information given is the postulate SSA (or ASS, the Donkey Theorem), but the two triangles that were created are clearly not congruent. We have two triangles with two sides and the non-included angle congruent, but the triangles are not congruent to each other.

Review

How to convert radians to degrees and degrees to radians.


Degrees into radians

360° = 2π. 180° = π.

Radians into degrees

Review

Basic Trigonometric Values

Review

Acute angle

An angle whose measure is less than 90o.

Obtuse angle

An angle whose measure is grater than 90o but less than 180o.

Right angle

An angle whose measure is 90o.

Reflex angle

An angle whose measure is greater than 180o but less than 360o.

A straight angle

An angle whose measure is 180o.


Central Angle
CentralAngle

A central angle is an angle ∠AOC with endpoints A and C located on a circle's circumference and vertex O located at the circle's center . A central angle in a circle determines an arc arcAC.

CentralInscribedAngle

For an inscribed angle ∠ABC and central angle ∠AOC with the same endpoints,

 ∠AOC=2∠ABC