Trigonometry

First, they contact Darth Vador. When he arrives, they locate the secret banana together and attach a string to it. They throw it into the ocean, and use trigonometry to locate dolphins. Then the fly off into space, and defeat Luke Skywalker.

The basic functions are sine, cosine, tangent, cosecant, secant and cotangent. In addition, there are their inverses, whose full names use the prefix "arc" [arcsine, arc cosine, etc] but are more often written as sin-1, cos-1 and so on.

the person who did this was ben Franklin

I am trying to find the answers to this....and no the answer is not ben Franklin...sorry

I agree it is not ben Franklin...good try though

They are classified as: scalene, obtuse, right angle. equilateral and isosceles

They mean that the expression to the left of the sign is greater than or less than (as appropriate) the expression to the right of the sign.

Assume the angle u takes place in Quadrant IV.

Let u = arctan(-12). Then, tan(u) = -12.

By the Pythagorean identity, we obtain:

sec(u) = √(1 + tan²(u))

= √(1 + (-12)²)

= √145

Since secant is the inverse of cosine, we have:

cos(u) = 1/√145

Therefore:

sin(u) = -√(1 - cos²(u))

= -√(1 - 1/145)

= -12/√145

Otherwise, if the angle takes place in Quadrant II, then sin(u) = 12/√145

That depends on your profession. If you are a math teacher, then you might use a lot of Trig. If you are an engineer, working with forces on any object from different directions, then you would use trig. Electrical engineers use trig. Surveyors use trig.

Trigonometry is essential to the study of higher mathematics (calculus) and to the understanding of many scientific and engineering principles. Trigonometry and calculus can be used to model many shapes, motions, and functions in daily life.

Every triangle have 6 main parts: 3 sides and 3 angles. On a right triangle one of the angles has to be a right angle, meaning it has a 90 degree angle.

A part of a circle's circumference is an arc

One radian is about 57.3 degrees

Assuming that neither of the given sides is the hypotenuse, then if A is one of the acute angles, tan(A) = 19/63

So A = arctan(19/63) = 16.8 degrees. The other acute angle is 73.2 deg.

The hypotenuse is a side, not an angle. However, if you mean the angle across from the hypotenuse, it is always 90 degrees, or pi/2 radians because hypotenuses only exist in right triangles.

Quadrant I.

'Theta' is the eighth character in the Greek alphabet. It is not often used as a mathematical constant (such as 'Pi'), but rather as a variable. It is commonly, but not always, used to represent some arbitrary angle.

1/2 x length x width x height

Surveying the land. Laying out the position of the foundation of the building.

You cannot "solve" ungrouped data since ungrouped data is not a question. You can calculate the mean or the variance, standard deviation or skewness, or a whole range of other measures for ungrouped data. But you have not specified what.

sin(0) = 0, sin(90) = 1, sin(180) = 0, sin (270) = -1

cos(0) = 1, cos(90) = 0, cos(180) = -1, cos (270) = 0

tan(0) = 0, tan (180) = 0.

cosec(90) = 1, cosec(270) = -1

sec(0) = 1, sec(180) = -1

cot(90)= 0, cot(270) = 0

The rest of them:

tan(90), tan (270)

cosec(0), cosec(180)

sec(90), sec(270)

cot(0), cot(180)

are not defined since they entail division by zero.

Top left corner

It is a measure of an area, approximately equal to 492 sq metres.

0=45 degrees?

We stat with the law of cosines, which we can assume to be true:

* c2 = a2 + b2 - 2ab*cos(C) Then rearrange it:

* cos(C) = a2 + b2 - c2/2ab Use the identity sin(x)=SQRT(1-cos(x))

* sin(C) = SQRT( 1 - (a2 + b2 - c2/2ab)2) Use the operator A = 1/2ab*sin(C) where A is area. Also, set one equal to 4a2b2 and factor it out.

* 2A/ab = SQRT(4a2b2 - (a2 + b2 - c2)2)/2ab ab's cancel, and the term inside the square root is the difference of two squares.

* A = 1/4*SQRT((2ab - (a2 + b2 - c2))(2ab + (a2 + b2 - c2))) when the two groups are simplified, the can be factored in binomial squares.

* A = 1/4*SQRT((c2 - (a - b)2)((a + b)2 - c2) Once again, we have differences of squares.

* A = 1/4*SQRT((c - (a - b))(c + (a - b))((a + b) - c)((a + b + c)) Simplify.

* A = 1/4*SQRT((c + b - a)(c + a - b)(a + b - c)(a + b +c)) Here comes the tricky part. We have four parts here. Three have two terms positive and one negative. Having a + b + c is like having the P. If we have a + b - c, that is like saying P - 2c, right? So to make it even easier, we can call s, the semi-perimeter, P/2. Then we can say a + b - c is 2s - 2c, or 2(s - c). We can apply that to all parts except the last one, which is just 2s.

* A = 1/4*SQRT(2(s - a)*2(s - b)*2(s - c)*2s) The two's can multiply together to 16 and come out of the root, canceling with the 1/4. we are left with good old Heron's formula.

* A = SQRT(s(s - a)(s - b)(s - c))

It's possible that either the angles or sides are labeled according to length or size.

Let's look at right triangles for a moment. In any right triangle, the hypotenuse is the side opposite the right angle. There exist three ratios (and their inverses) as regards the length of the sides of the right triangle. These are opposite/hypotenuse (called the sine function), adjacent/hypotenuse (called the cosine function), and opposite/adjacent (called the tangent function). The inverse of the sine is the cosecant, the inverse of the cosine is the secant, and the inverse of the tangent is the cotangent. The abbreviations for these functions are, sin, cos, tan, csc, sec and cot, respectively.

What is underneath this idea is that for any (every!) right triangle, there is a fundamental relationship or ratio between the lengths of the sides for all triangles with the same angles. For instance, if we have a triangle with interior angles of 30 and 60 degrees (in addition to the right angle), regardless of what size it is, the ratio of the lengths of the sides is always the same. And the trigonometric functions express the ratios of the lengths of the sides.