How do you plot a set of points to find a 52 degree angle in a right triangle?
494152
From the point where your 52 degree angle is going to be, measure 1-9/16" over and 2" up.  The angle nearest your 1-9/16 measurement will be pretty close to 52 degrees (actually about 52.00126756, but in making jigs and other shop fixtures, I think 1-9/16" over and 2" up are pretty close).
892974
a right triangle is called a right triangle because there is a 90 degree angle in it.
889937
A triangle has 180 degrees.  a right angle is 90 degrees.  a triangle has 3 angles.
1030296
for it to be a triangle the angles have to measure to 180, so if you have a 90 degree andlge off one, you could subtract 90 from that, take the remaining 90 and work with that, since you know your need a 1:1 of the legs , then you know you need an equal angle.. angle has to equal 90/2... and thats where i get stumped...\n      gl
723227
<R= 52* 180-83-45=52. because 3 <s of a triangle is 180.\n\nLaw of Sins:\nSin A / a = Sin B / b = Sin C/ c\n\n(sin 52)/ 18 = (sin 45)/ q\ncross multiply\n[ 18 x (sin 45) ] / (sin 52) = q\nq= 16.1 so about 16\n\n(sin 52)/ 18 = (sin 83)/ p\ncross multiply\n[ 18(sin 83) ] / (sin 52) = p\np=22.67 so about 23\n\nthere!\n<R = 52*\nq= 16\np= 23
733210
You basically have 2 options. Use a protractor and measure them (The internal angles total 180 degrees).\n\nThe other option is using the trig functions (Sin, cos and tan), however you must have some info on the triangle eg. length of sides to apply any of these.
793698
Start by drawing a picture of this.  Draw a right triangle.  I put the right angle at the bottom left.  Then draw an arbitrary line from the upper left vertex down to the base.  This separtes the large right triangle into a smaller right triangle and what I will call the "outer triangle."  This gives you the 1000 Jim moved; it's to the right of where the line intersects the base.  I called this leg a.  I drew in the angles I was given, and labeled all of the line segments.  The hypotenuse of the large right triangle is b, the hypotenuse of the smaller right triangle (also the only unlabeled leg of the outer triangle) is c, and the vertical leg of the right triangle is d.  d is the height of the escarpment.  I named the angles after the sides they face.  So the 10 degree angle is C, the 32.5 degree angle is D, and angle A is opposite leg a, and is an acute angle in the outer triangle.  Since I was given angle D, I could find its supplement, B, which is an obtuse angle in the outer triangle, opposite leg b.  B = 180 - D = 180 - 32.5 = 147.5.  Since the sum of angles in a triangle is 180, and I know C and B, I can also find A = 180 - B - C = 180 - 147.5 - 10 = 22.5.\n\nNow that I know every angle in the outer triangle and the length of one leg, a, I can use the law of sines.  The law of sines says that a / sin(A) = b / sin(B).  Rearranging, b = a*sin(B) / sin(A) = 1000*sin(147.5) / sin(22.5).  Now look at the larger right triangle.  You know the length of the hypotenuse, b, and you know the angle C opposite the unknown d (I called it C because it is also opposite leg c in the outer triangle.).  By trigonometry, d = b*sin(C) = b*sin(10), where we use b from above.
857791
Without a picture or more information, your question isn't answerable, but I'm guessing that angles x and a are parts of a right angle, in which case the answer would be 41 degrees (49 + x = 90).
715400
Angle of line from horizontal position.
1117196
use cosine law (because you don't know any of the angles) \n\n---- cosine law used when you know the measurement of an angle between 2 known sides or when there're no known angles at all\n\na^2= b^2 + c^2 - 2bc x cosA\n\nb^2= a^2 + c^2 - 2ac x cosB\n\nc^2= a^2 + b^2 - 2ab x cosC\n\nc is side opposite angle C, a the side opp angle A, b the side opp angle B.\n\nyou first have to draw the triangle\n\nafter drawing it, you will see that the side opp angle A is BC, and it is 7 units long (label the sides as a, b, or c if that helps you). what you do afterwards is put the values into the equation.\n\n(7)^2= (12)^2 +(10)^2 - 2(12)(10)cosA\n\nisolate the unknown value\n\n49 - 144 = 100 - 2(12)(10)cosA\n-95-100 = -2(12)(10)cosA\n-195 = -2(12)(10)cosA\n-195/(-2 x 12 x 10) = cosA\ncosA = 0.8125\n\nto find A, just press cos^-1 0.8125 on your calculator.\n\nangle A = cos^-1 0.8125\nangle A = 35.65908.... degrees\n\nrounded to nearest degree, angle A is 36 degrees\n\nhope that helped :)
607716
If the circles are all tangent to each other, then you can draw lines from center to center to center.  Thus, you'd have a triangle with sides 8, 12, and 10 (each side is formed by adding two radii).  So you can find the angles using the law of cosines:\n\nc^2 = a^2 + b^2 - 2abcos(c)\n8^2 = 12^2 + 10^2 - 2(12)(10)cos(c)\n64 = 144 + 100 - 240cos(c)\n240cos(c) = 180\ncos(c) = 3/4\nc ~ 41 deg\n\nAnd then just change up numbers to find the other angles:\n\nb^2 = a^2 + c^2 - 2accos(b)\n10^2 = 12^2 + 8^2 - 2(12)(8)cos(b)\n100 = 144 + 64 - 192cos(b)\n192cos(b) = 108\ncos(b) = 9/16\nb ~ 56 deg\n\na^2 = b^2 + c^2 - 2bccos(a)\n12^2 = 10^2 + 8^2 - 2(10)(8)cos(a)\n144 = 100 + 64 - 160cos(a)\n160cos(a) = 20\ncos(a) = 1/8\na ~ 83 deg\n\nAnd this checks out since 41 + 56 + 83 = 180.
885675
Check out spherical trigonometry.  The sum of the angles in a spherical triangle can be from 180 to 540 degrees!\n\nhttp://mathworld.wolfram.com/SphericalTriangle.html
701162
Draw the triangle on an x-y plot with the right angle at 0,0 and the other two points at (12,0) and (0,5). First write an expression for the hypotenuse line.  The y-intercept is B = 5 and the slope is:\nm = rise/run\nm = -5/12\n\nthe equation of the line would then be:\ny = mx + B\ny = (-5/12)x + 5\n\nnow calculate the area of a rectangle:\nA = x*y\nA = x*((-5/12)x + 5)\nA = (-5/12)x^2 + 5x\n\nto get the maximum you need to take the derivative and set it to 0:\ndA/dx = 2*(-5/12)x + 5\n0 = (-5/6)x + 5\n(5/6)x = 5\nx = 6\ny = (-5/12)x + 5\ny = (-30/12) + 5 = 2.5\n\nYou should be able to do part b on your own now.\n\nFor part c just replace the numbers with variables:\na will be the length and b will be the height\n\nThe y-intercept is B = b and the slope is:\nm = rise/run\nm = -b/a\n\ny = (-b/a)x + b\n\nnow calculate the area of a rectangle:\nA = x*y\nA = x*((-b/a)x + b)\nA = (-b/a)x^2 + bx\n\nto get the maximum you need to take the derivative and set it to 0.  Treat b and a just as if they were actual numbers:\ndA/dx = 2*(-b/a)x + b\n0 = (-2b/a)x + b\n(2b/a)x = b\nx = a/2\ny = (-b/a)x + b\ny = (-b/a)(a/2) + b\ny = (-b/2) + b\ny = b/2
882048
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
454825
We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
777719
Mark you first two corners 20 m apart.\nThe third marker must be 20 m away from one corner and the other must be 20*sqrt(2) = 28.284 meters from the other corner (The length of the diagonal of a square is sqrt(2) * the side, so that distance will ensure that you have right angles.)\n\nSo tie a length of rope (no stretch), to each of the first two markers.  Cut the ends at 20m and 28.284 m respectively.  Hold the two ends together and find the spot where both ropes are taut.  That's where your third marker goes.\n\nAgain, your last marker should be 28.284m from its opposite.
803546
Use the pythagorean Therym(?)\nThat is, well, here is an example:\nIf one leg of a right triangle is 6 units, and the other leg is 8 units, what is the legnth of the hypotenuse?\nthat is:\nleg 1=a, leg 2=b, and the hypotenuse = c\nThe hypotenuse always = c, it is the side parallel to the right angle of the triangle.\nthe pythagorean therym(?) is a^2(a squared)+b^2(b squared)=c^2(c squared)\na=6, b=8, and c=?\n6^2+8^2=c^2\n36 + 64= c^2\n100= c^2\nc=10^2\nso the missing side would b 10 units, just use that example for your problem, and remember, if you are trying to find the legnth of a leg, just plug in the one leg length that you have (this would be "a") and the hypotenuse(this would be "c") and subtract the square of "a" from "c". find the square root, and vua la, there you have it. hope that helps