Making Clinometer

This Project is contributed by Mohammad Wasif X F

Introduction
WHAT IS TRIGONOMETRY?

THE WORD TRIGONOMETRY IS DERIVED FROM THE GREEK WORDS:
I)TRIGONO AND 2)METRON.THE WORD TRIGONON MEANS A TRIANGLE AND THE
WORD METRON MEANS A MEASURE.HENCE TRIGONOMETRY MEANS THE SCIENCE OF MEASURING TRIANGLES.
Line of sight:- is the line drawn from the eye of an observer to the point in the object viewed by the observer.
Angle of depression:-of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level.
Angle of elevation:-of a point viewed is the angle formed by the line of sight with the horizontal when the point being viewed above the horizontal level.
A clinometer is an instrument that measures vertical slope, usually the angle between the ground or the observer and a tall object, such as a tree or ! building.

Aim :-TO MAKE A CLINOMETER
Material Required :
1-Protactor
2-Straw
3-Stone
4-Thread


















Procedure
Step 1 Get a protractor with one straight edge (a 180 degree protractor).




















Step 2
Tape a straw along the straight edge of the protractor.



















Step 3.
Tie a string through the small hole on the straight edge that is directly across from the 0 degree mark on the protractor. This may also be labelled as 90 degrees. If your protractor does not have a small hole here, or if the hole is not situated correctly (this is a common problem with some chea! p protractors), tape or glue the string to the protractor at this mark. Make sure the string dangles a few inches below the protractor.








Step 4
Attach a washer or fishing weight to the dangling end of the string.







!











Step 5
Sight the top of a tall object through the straw.



















Step 6
Note the number where the string crosses. Subtract this number from 90 to determine the angle of elevation between your eye and the top of the object you are sighting

In the above case I measured the angle of elevation and observed that the angle is 36 degrees. According to the above rule I will have to subtract the measured angle with 90 degrees. So the angle of elevation is 90 degrees -36 degrees = 54 degrees.

Conclu! sion:
From the above discussion we have seen that we can easily find out the Heights and Distances without any actual measurement. We used only clinometers to measure the angle of elevation & angle of depression. with the help of trigonometric formulae's we can easily find out the height and distances.
Astronomers can easily find the distance of stars from earth with help of trigonometric formulae's.

Resources:

Planet infinity.
Mathematics Encyclopedia.

I would like to congratulate Mohammad on behalf of Planet Infinity for doing such a wonderful job.

Heights and Distances

Plates found under crop circles... interesting

Alien artifacts discovered underneath crop circles

Circles and Circle Properties

Drawing WPF Curves with Arrow heads

Having drawn the shapes, the curves and found their intersection point, I finally needed to draw an arrow head. In order to draw the arrow at the right angle we need to know the tangent of the Bezier curve at the intersection point. Since that is beyond my maths capability I chose to find the intersection point between the line and an imaginary, slightly larger, rectangle that surrounds the target shape. The difference between the two intersection points provides the required angle for the arrow head.

private static void DrawArrowHead(Canvas canvas, PathGeometry linePath, Rect shapeRect, Color color)
{
   // Get the intersection point of the imaginary, slightly
   // larger rectangle that surrounds the targer shape.
   Rect!   outerRect = new Rect(shapeRect.Left - 10, shapeRect.Top - 10, shapeRect.Width + 20, shapeRect.Height + 20);

   RectangleGeometry shapeGeometry = new RectangleGeometry(shapeRect);
   Point[] intersectPoints = GetIntersectionPoints(linePath, shapeGeometry);


   double innerLeft = intersectPoints[0].X;
   double innerTop = intersectPoints[0].Y;


   shapeGeometry = new RectangleGeometry(outerRect);
   intersectPoints = GetIntersectionPoints(linePath, shapeGeometry);


   double outerLeft = intersectPoints[0].X;
   double outerTop = intersectPoints[0].Y;


   Polygon arrowHead = new Polygon();
   arrowHead.Points = new PointCollection();
   arrowHead.Points.Add(new Point(innerLeft, innerTop));
   arrowHead.Points.Add(new Point(innerLeft + 10, innerTop + 5));
   arrowHead.Points.Add(new Point(innerLeft + 10, innerTop - 5));
   arrowHead.Points.Add(new Point(inner!  Left, innerTop));
   arrowHead.Stroke = new SolidColorBru!  sh(color );
   arrowHead.Fill = new SolidColorBrush(color);


   // The differences between the intersection points on
   // the inner and outer shapes gives us the base and
   // perpendicular of the right-angled triangle
   double baseSize = innerLeft - outerLeft;
   double perpSize = innerTop - outerTop;
   // Calculate the angle in degrees using ATan
   double angle = Math.Atan(perpSize / baseSize) * 180 / Math.PI;


   // Rotate another 180 degrees for lines in the 3rd & 4th quadrants
   if (baseSize >= 0) angle += 180;


   // Apply the rotation to the arrow head
   RotateTransform rt = new RotateTransform(angle, innerLeft, innerTop);
   arrowHead.RenderTransform = rt;


   // Arrow heads are drawn over the lines but
   // under the shapes
   Canvas.SetZIndex(arrowHead, (int)Layer.Arrow);


   canvas.Children.Add(arrowHead);
}
 
 
Angle of intersection between curves
 

Work Problem 5

The following was a question a anonymous visitor asked: Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?



Work problem 5 solution here

pre online agebra answers

Math and Algebra Online Homework Help from Tutor Vista

For some genius mathematic probably is a number game and a challenge that will be easily to solve. But for mostly students and people, mathematic is a serious problem. Some students have to get additional lesson at some course at night just to improve their mathematic score at the school.

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Online math help

Here I found another interesting a site that provide online math help. If you are looking for help on your math lesson, Guaranteach might suit your need.  They have more than 20000 lessons that you can access. The interesting part is, this lesson is in video format. Another interesting fact is that they provide some interesting method to detect your learning style. This of course will be great advantage for you, you can maximize your learning process by learning on your own style. Don't worry about the quality of the tutorial, the lessons is teacher approved and more importantly, you can access it whenever and wherever you want it. Don't ruin your math learning because you're shy to ask a question in the class. With those thousands online tutoring that Guaranteach has, I ! believe you will overcome your difficulties on math lesson.

Did they provide video only? No. This is getting more interesting about Guaranteach service. Like a normal class-teaching, they also has quizzes and progress report. Isn't it great? Now, you can measure your learning progress just like normal without exposing your result to your friend (well, honestly, I, sometimes got bad mark on math that I really really shy on it -- being laughed by my classmates). Before applying for their service, you may try it for free. They provide up to 5 lessons videos that you can watch, so you know how's their lessons quality. The registration is quick and no need for credit card (obviously, most of us -- student -- didn't have any credit card yet). Go try them out!

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Precalculus Formulas

Here you can find formulas for a Precalculus Course.

1.1 Trigonometric functions
For the trigonometric ratios for a point p on the unit circle holds:
cos(φ) = x_p , sin(φ) = y_p , tan(φ) = y_p/x_p
sin² (x) + cos² (x) = 1 and cos⁻² (x) = 1 + tan² (x).
cos(a ± b) = cos(a) cos(b) ∓ sin(a) sin(b) ,
sin(a ± b) = sin(a) cos(b) ± cos(a)sin(b)
tan(a ± b) =( tan(a) ± tan(b) )/( 1 ∓ tan(a) tan(b) )
The sum formulas are:
sin(p) + sin(q) = 2 sin( 1 / 2 (p + q)) cos( 1 / 2 (p − q))
sin(p) − sin(q) = 2 cos( 1 / 2 (p + q)) sin( 1 / 2 (p − q))
cos(p) + cos(q) = 2 cos( 1 / 2 (p + q)) cos( 1 / 2 (p − q))
cos(p) − cos(q) = −2 sin( 1 / 2 (p + q)) sin( 1 / 2 (p − q))
From these equations can be derived that
2 cos2 (x) = 1 + cos(2x) , 2 sin2 (x) = 1 ! 722; cos(2x)
sin(Ï€ − x) = sin(x) , cos(Ï€ − x) = − cos(x)
sin( 1 / 2 Ï€ − x) = cos(x) , cos( 1 / 2 Ï€ − x) = sin(x)
Conclusions from equalities:
sin(x) = sin(a)
⇒ x = a ± 2kÏ€ or x = (Ï€ − a) ± 2kÏ€, k ∈ N
cos(x) = cos(a)
⇒ x = a ± 2kÏ€ or x = −a ± 2kÏ€
tan(x) = tan(a)
⇒ x = a ± kÏ€ and x =/= Ï€ / 2 ± kÏ€
The following relations exist between the inverse trigonometric functions:
arctan(x) = arcsin(x/ √x² + 1)= arccos(1/√ x² + 1)

sin(arccos(x)) = √ 1 − x²

precalculus problems solved