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Posted on Jan 02, 2017

I find the easiest way to solve these is to sketch them first (I'm a visual learner;) We get a nice right-angled triangle, with the right-angle at B. The formula for the area of a triangle is 1/2 * base* height or (base * height)/2.

We can use BC or AB as the base.

If we use BC as the base, the length is 9-4 or 5. The height is 6-2 or 4.

We can now but the base and the height in the formula to figure out the area.

Good luck.

Paul

We can use BC or AB as the base.

If we use BC as the base, the length is 9-4 or 5. The height is 6-2 or 4.

We can now but the base and the height in the formula to figure out the area.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

There are two adjustment points (screws). For example left headlight: you will see one point near the end of left wing, at bottom side through a wing hole. This is for vertical plane headlight beam adjustment (up-down). Another one point is against water radiator side. This is for horizontal plane headlight beam adjustment (left-right).

on Dec 18, 2009 | Audi A4 Cars & Trucks

The area of a triangle is 1/2 times base times height. A sketch of the triangle in the coordinate plane will determine how easy or hard this will be to be. From the sketch, you will see that this is a right-angled triangle with B being the right-angle. This makes it easier because we can easily determine the base and the height to use in the formula.

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

The area of a triangle is 1/2 times base times height. A sketch of the triangle in the coordinate plane will determine how easy or hard this will be to be. From the sketch, you will see that this is a right-angled triangle with B being the right-angle. This makes it easier because we can easily determine the base and the height to use in the formula.

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

In which language? This is really more a mathematical problem than a programming/compilation one. Try walking the circumference of the circle, testing for a point along each of the triangle's segments. If only one such point can be found, or a series of consecutive points (since software "circles" are never true circles, they're made up of short straight segments), then you have a tangent.

You can save a lot of time and math by making coarse estimates first - for example, don't bother testing for tangentiality for any circumferential segment that's fully above, to the left or right of, or below, all three points of the triangle - you CANNOT have a tangent in those areas.

An alternate approach is... walk the triangle instead of walking the circle. Test for a single circle crossing point. Again, you can save time by checking each segment first, and walking only segments that actually cross the circle's circumference. If no part of the circle can be found mathematically between any pair of vertices, then discard that segment outright.

You can save a lot of time and math by making coarse estimates first - for example, don't bother testing for tangentiality for any circumferential segment that's fully above, to the left or right of, or below, all three points of the triangle - you CANNOT have a tangent in those areas.

An alternate approach is... walk the triangle instead of walking the circle. Test for a single circle crossing point. Again, you can save time by checking each segment first, and walking only segments that actually cross the circle's circumference. If no part of the circle can be found mathematically between any pair of vertices, then discard that segment outright.

Mar 10, 2014 | Computers & Internet

You need 5 isosceles triangles with a base of 30 mm. To get the height of the triangles (perpendicular to the 30 mm bases) you need to calculate the apothem of the pentagon (assumed to be regular).

If you cut the pyramid by a plane passing through its apex, the center of the base-pentagon,and the midpoint of one side, the plane figure created by the three points (apex, center, and midpoint) is a right triangle. The legs are the apothem, and the altitude of the pyramid. The hypotenuse is the slant height of the pyramid, and is thus the height of the triangles in the development. pyramid (60 mm) form.

Use the Pythagorean Theorem to find that slant height.

**s^2=a^2+h^2**.

If you cut the pyramid by a plane passing through its apex, the center of the base-pentagon,and the midpoint of one side, the plane figure created by the three points (apex, center, and midpoint) is a right triangle. The legs are the apothem, and the altitude of the pyramid. The hypotenuse is the slant height of the pyramid, and is thus the height of the triangles in the development. pyramid (60 mm) form.

Use the Pythagorean Theorem to find that slant height.

Nov 11, 2013 | Computers & Internet

There are two adjustment points (screws). For example left headlight: you will see one point near the end of left wing, at bottom side through a wing hole. This is for vertical plane headlight beam adjustment (up-down). Another one point is against water radiator side. This is for horizontal plane headlight beam adjustment (left-right).

Nov 19, 2009 | 2004 Audi A4

There are two adjustment points (screws). For example left headlight: you will see one point near the end of left wing, at bottom side through a wing hole. This is for vertical plane headlight beam adjustment (up-down). Another one point is against water radiator side. This is for horizontal plane headlight beam adjustment (left-right).

Please Rate my Response! Thanks!

Please Rate my Response! Thanks!

Jul 13, 2009 | 2003 Audi A4 Cabriolet

Hi again,

Ok, I finally got it licked and why you're having the problems and it's fairly simple. It's all based on 2 things but most of all, the order in which you create your planes or grids.

Because the main part of the image is on the front face that is what you want as your Parent face so get that as accurate as you can to start off with. Hold the 'x' key to temp zoom in and get the position as correct as you can.

Once you've created you're primary ctrl-click(cmd-click on mac) the mid top point and pull it out then the same for the mid left point.

Now if it still has the problem when you paste in the image and drag it onto the primary face then undo the paste. We need to adjust the secondary planes. Select the top plane and in the angle parameter (which should show 270°) change it to 269°. Then change the left planes angle to 89°. Try it now and it should work fine. Then put it back to 270° and 90° again.

Let me know how you get on.

All the best

Sean

PiDigital.co.uk

Ok, I finally got it licked and why you're having the problems and it's fairly simple. It's all based on 2 things but most of all, the order in which you create your planes or grids.

Because the main part of the image is on the front face that is what you want as your Parent face so get that as accurate as you can to start off with. Hold the 'x' key to temp zoom in and get the position as correct as you can.

Once you've created you're primary ctrl-click(cmd-click on mac) the mid top point and pull it out then the same for the mid left point.

Now if it still has the problem when you paste in the image and drag it onto the primary face then undo the paste. We need to adjust the secondary planes. Select the top plane and in the angle parameter (which should show 270°) change it to 269°. Then change the left planes angle to 89°. Try it now and it should work fine. Then put it back to 270° and 90° again.

Let me know how you get on.

All the best

Sean

PiDigital.co.uk

Dec 28, 2008 | Adobe Photoshop CS3 10

Debs,

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.

Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311

- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:

square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.

Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311

- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:

square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

Dec 19, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

Aug 29, 2017 | Canon Cameras

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