Trigonometry

Yes.

Rather a convoluted way of saying it, but it is true.

It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines.

sin x / (1 - cos x) = csc x + cot x

sin x / (1 - cos x) = 1 / sin x + cos x / sin x

Now it should be easy to do some simplifications:

sin x / (1 - cos x) = (1 + cos x) / sin x

Multiply both sides by 1 + cos x:

sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x

sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x

sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x

sin x (1 + cos x) / sin x = (1 + cos x)2

1 + cos x = (1 + cos x)2

1 = 1 + cos x

cos x = 0

So, cos x can be pi/2, 3 pi / 2, etc.

In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.

There is no largest Pythagorean triple since there's infinite amount of them. But if you're looking for one quite big, I took a few minutes for you and wrote a program that computes them (and btw is still computing them). Right now the largest one the function returned is 77893200, 128189952, 150000048. Note that you can multiply all three with any same natural number larger than one (2,3,4,...) and you'll get a Pythagorean triple larger than mine.

No.

It is a plane figure bounded by four straight lines, two of which are parallel.

Yes.

Using Pythagorean's theorem: 202+212 = 400+441 = 841 = 292 = 841

0.525321989

Round the base angle to 70 degrees and use the sine ratio:

30*sine 70 degrees = 28.19077862 feet

Height of ladder from the ground = 28 feet to 2 s.f.

45 degrees (+/- 180k degrees for any integer k)

or pi/4 radians (+/- pi*k radians for any integer k).

The ladder forms a right angle with the building: the ground and the building forming the right angle and the ladder forming the hypotenuse. If the length of the ladder is L metres, then

sin(49) = 12/L

So L = 12/sin(49) = 15.9 = 16 metres.

This describes a right triangle. This triangle has a base (X ) of 3 ft, a opposite side ( Y) of 9 ft. So, you are looking for the hypothenuse. Use the Pythagoreum theory.

In this case. Your ladder length is called H.

H^2 = X^2 + Y^2

H = sqrt X^2 + Y^2

50.3625 in DMS form is 50.21.45

50 is the D part.

Subtract 50 to get 0.3625. Multiply by 60 to get 21.75. 21 is the M part.

Subtract 21 to get 0.75. Multiply by 60 to get 45. 45 is the S part.

Approx 3440 sq metres.

Use a protractor.

Let 'theta' = A [as 'A' is easier to type]

sec A - 1/(sec A)

= 1/(cos A) - cos A

= (1 - cos^2 A)/(cos A)

= (sin^2 A)/(cos A)

= (tan A)*(sin A)

Then you can swap back the 'A' with theta

sin(x) = tan(x) when x equal 0

You can use the inverse of sin when you want to solve an equation where x is the angle you're trying to find.

Say sin(x)=32/50

Since you can't plug "x" into your calculator, use the arc sin (represented on your calculator by sin -1) on both sides to get rid of the sin. This is how it would plug into your calculator:

sin-1 (32/50)

Whatever the answer is would be what "x" equals.

Use the tangent ratio:

tan = 22.5/34 = 45/68

tan-1(45/68) = 33.49518467 degrees

Angle of elevation = 33o29'42.66''

Approx 2.6131

y=sin x

y=cos x cos x

sin x = cos^2 x

sin x = 1-sin^2 x

sin x -1 + sin^2 x = 0

sin^2 x + sin x -1 = 0

Let y=sin x

y^2+y-1 = 0

This equation is of form ay^2+by+c=0

a = 1 b = 1 c = -1

y=[-b+/-sqrt(b^2-4ac)]/2a]

y=[-1 +/-sqrt(1^2-4(1)(-1)]/(2)(1)

discriminant is b^2-4ac =5

y=[-1 +√(5)] / 2

y=[-1 -√(5)] / 2

sin x = [-1 +√(5)] / 2

x = sin^-1 [-1 +√(5)] / 2] = 0.6662394 radians

x = sin^-1 [-1 -√(5)] / 2] = sin^-1 (-1.618) -- has no solution

When x = 0.6662394 radians, sin x and cos x times cos x are equal.

Leaving aside the fuzziness caused by atmospheric diffraction the curve (called the Terminator!) is really a straight line running North-South. North and South in this context are "poles" perpendicular to the plane of the earth's orbit around the sun and should not be confused with the geographic or magnetic N and S.

Unfortunately for map makers, the earth is approximately spherical and so they need to use some form of projection to represent the earth on a 2-dimensional (flat) surface. This process distorts the geography of the earth. For example, the most commonly used projection, the Mercator projection, greatly exaggerates the size of the polar regions with this exaggeration reducing as one moves towards the equator. Other projections introduce other distortions. Since the earth's axis is tilted, the NS for the Terminator are different from the geographic NS and as a result the terminator appears as a curve. Its exact shape is determined by the projection used even though in reality it is a straight line.

I assume. Since theta is a variable, standing for the measure of any angle.

BEARING

1. A, B and C are three ships. The bearing of A from B is 045º. The bearing of C from A is 135º. If AB= 8km and AC= 6km, what is the bearing of B from C?

2. A helicopter pilot flies in the direction 147 degrees and lands when she is 12 km south of her starting point. How far did she fly?

Tanx can be written as tan0.5x by dividing it by 2.

tanx1/2=tan0.5x

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I doubt that you can, since the tangent of the whole angle is a function of the tangent of the half-angle and of the secant of the whole angle. Please see the link.

An obtuse angle.