Trigonometry

In a problem which can be represented by a right angled triangle, the value of the cosine of an angle can be found by dividing the length of the side adjacent to the angle with the length of the hypotenuse, so if the shorter side was 16 units long and the hypotenuse was 25 units long the cosine of the angle between those two lines would be 16/25 = 64/100 = 0.064;these cosines can easily be found on even cheap hand calculators; it is more difficult and rather obsolete to look for them in a book of mathematical tables

0.813

Imagine a random triangle ABC. It will make it easier if you draw it with angle C at the top. The opposite side of angle A is labelled a, the opposite side of angle B is labelled b and the opposite side of angle C is labelled c. Draw the altitude (height) from angle C so that it is perpendicular (at 90 degrees) to side c.

Looking at this triangle, find expressions for the sines of angles A and B:

sinA = h/b

sinB = h/a

Rearrange these two equations in terms of h:

h = bsinA

h = asinB

As h = h, these equations can be set equal to each other and simplified to find the sine rule:

bsinA = asinB

sinA/a = sinB/b

If you expand on this way of working, you can also find that sinA/a = sinB/b = sinC/c. You have now proven the sine rule for all triangles!

This question doesn't mean anything. Explain more thoroughly.

Hipparchus is the father of trigonometry.

If the polygon (not ploygon!) has n vertices, the sum of the angles is (n-2)*pi radians or (n-2)*180 degrees.

(30+360*k) and (150+360*k) degrees where k is any integer.

Assume the first term is cos-squared(theta) rather than cos(2*theta).

6cos2(t) + 5cos(t) - 4 = 0

[6cos2(t) + 8cos(t) - 3 cos(t) - 4] = 0

[3cos(t) + 4]*[2cos(t) - 1] = 0

which gives

3cos(t) = -4 so that cos(t) = -4/3

or

2cos(t) = 1 so that cos(t) = 1/2

The first of these is clearly not a possible solution, whereas the second can hve one or more solutions, depending on the domain - which is not specified in the question.

If the initial assumption is incorrect and the first term WAS 6cos(2*theta) then using the formula for double angles:

cos(2(t) = cos2(t) - sin2(t) = cos2(t) - [1 - cos2(t)] = 2cos2(t) - 1

and so, the equation becomes

6*[2cos2(t) - 1] + 5cos(t) - 4 = 0

or 12cos2(t) + 5cos(t) - 10 = 0

Solve this quadratic equation for cos(t) and then find the values of t (or theta) within the domain specified.

It is the distance between opposite corners and it can be worked by using Pythagoras' theorem.

It is an obtuse angle.

iota

cos = adjacent / hypotenuse

2.9

The Greek letter theta

5 mile = 8 km

5/8 of 65 = 40.625

Answer: 40.625 mph

d theta divided by 4.28 is not an equation or inequality: it is an expression. An expression cannot be solved.

Fourth.

It means that the sides are of equal lengths and that the interior angles are of equal sizes

• 277323 dividede by 2291

When a metal can pounded into shape it is called malleable. Malleability is a physical property associated with metals where they can be hammered or pressed into a shape.

Because it is a branch of mathematics that is concerned with measurements of angles and sides of triangles, and following from that, areas and other characteristics.

It is a right angle triangle and its largest angle is 90 degrees

I just called my cardiologist's office for an answer to this one (I'm scheduled for a nuclear stress test that uses the Bruce protocol). Here's what the person who will administer my test told me:

It's the percentage of 90 degrees, 0% being no incline, 100% being straight up, perpendicular to the floor. The test begins at 1.7 mph with an incline of 10%, or 9 degrees. The incline is raised every three minutes by 2%, or 1.8 degrees. So at Level 2, the incline is 10.8 degrees, Level 3 12.6 degrees, Level 4 15.2 degrees, etc. -Joe

19

If your asking how, if you put your finger on 10 and 21, move them towards the middle until they hit the same place, if they appear next to each other then the number in between it is the median and if there is many numbers, do the first way with those numbers and there you have it, pretty simple but maybe a bit hard to understand. if you do not understand my way, ask your math teacher and she will tell you how.

Arctan (49.22) = 88.83608° or 1.55048 radians.