ok, we are doing inscribed triangles, you have a circle with a rectangle in it. the points of the rectangle connect with the circle points(AB on the top and DC on the bottom)in the rectangle at angle a it's x and angle b is y , angle d = 102 degrees and c= 100 degrees. i know that y=78 degrees and x equals 80 degrees b/c the opposite angles equal 180 degrees. how do you find the arc of BCD? \nCan u understand how if worded it, draw it out please, thanks a lot!!!!!!!!!!!
135600
160. double X. because X is an inscribed angle. sry it took so long, i had to draw it =)
251348
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
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angle cdb = ancle cbd   (since traingle cbd is an isosceles)\n\nangle bcd =  180  -  (2*65)   =   50\n\nangle  acb =   180 - 68 - 30 =  82\n\n\n\nUsing sine rule,  (considering traingle  abc)\n\n(ab/sin 82) =  (bc / sin 68)\n\ngiven: ab= 5  ,  therefore  bc = 4.68   (2 decimal places)\n\n\n\n\nusing sine rule again,   (for triangle bcd)\n\n(bd/sin 50) =  (bc / sin 65)\n\nbc= 4.68   ,  therefore  bd = 3.96
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Let the three angles be X,Y Z with Y = 30 \n\nSince the sum is 180 we have:\nX + 30 + Z = 180 ......(1)\n\nand we are given\n X = 2( 30 + Z).........(2)  \n\nSubstituting we have (60 +2Z) + 30 +Z = 180\n\nSolving for Z we get  Z = 30 degree \naccordingly X = 180 -30 - 30 =120  \n\nAns: X= 120, Y =30 (given) Z= 30
376123
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.
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Since there are four quadrant that make up the 360degrees. Draw the two axis x and y. Assume the top right part as quadrant 1, top left part as quadrant 2, bottom left as 3 and bottom right as 4. Now the calculation of angle starts from quadrant 1 in anticlockwise direction to make an angle of 360degrees. Now use "ASTC" rule i.e. assign 'A' to quadrant 1, 'S' to quadrant 2, 'T' to 3 and 'C to 4. \n\n'A' states that in this quadrant all i.e. sin, cos and tan have positive values.\n\n'S' states that in this quadrant sin have positive value and cos and tan have negative values.\n\n'T' states that in this quadrant tan have positive value and cos and sin have negative values.\n\n'C' states that in this quadrant cos have positive value and sin and tan have negative values.\n\nthat's it simple enough.
201089
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
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Hi. Don't you hate these? OK you have an unknown angle. Call it a. You have a complement, c, which is 90-a. c=3a-10. Any help?
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Your statement is incorrect.  It reads:\n\n"In the image below, there are right angles. How would I prove that they're right angles without a protractor in any triangle?"\n\nActually, it should read:\n\n"In the image below, there are some angles."\n\nAnything else is pure conjecture.  (That means a guess.)\n\nIf you would like us to say something about those angles, you need to tell us how the figure was constructed.  After you tell us that, and ask your question more clearly, your answer will be obtained in short order.\n\nedit...\n\nYou have added some text in your extra text file that doesn't help at all:\n\n"I know that these green points are right angles..."\n\nSo you think a point can also be an angle.  It will be a long time before you get someone (who knows and understands geometry) to agree with that.\n\nIf you start over and explain what you want to know, in correct terms, you may have a chance at getting an answer.
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Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
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ok: i hope you sketched this out: forget for a minute the solid's diagonal is 26cm. let's use the ratios as actual units. so the one side with a 3*4 rectangle has a 5unit diagonal (from Pythagoras).\nnow-picture the right triangle with the two adjacent sides to the right angle, one side is 5 and one is 12 (that's the solid's diagonal). using OUR units for now, THAT diagonal is 13units (Pythagorean again). luckily, 26 is 2 * 13, so hopefully the measurements are 6,8,24, would such a solid yield a diagonal of 26?
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If they're tangent to each other that means they all touch at one point. That means you're given the lengths: 85 cm, 100 cm and 115 cm.\n\nFrom here, you can use the cosine law to figure out what each angle is.\n\na^2 = b^2 + c^2 - 2bc cosine A.
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Circles. You can inscribe an angle in a circle. Choose any point on the circle and let that be the vertex of your angle. Have the sides of the angle be chords of the circle. The degrees of arc subtended by the angle will be twice the number of degrees in the angle.\n\nAn inscribed angle of 90 degrees subtends a semi-circle.
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PART (a)\nUsing Pythagoras’ Theorem in triangle ABC,\nAB^2 + BC^2 = AC^2 = 85 [since AC = sqrt(85)]\nBoth sides AB and BC has to be positive integers.\nBy trial we take AB = 6, BC = 7 [since 6^2 + 7^2 = 36 + 49 = 85]\n\nAgain, using Pythagoras’ Theorem in triangle CDE,\nCD^2 + DE^2 = CE^2 = 85 [since CE = AC = sqrt(85)]\nBoth sides CD and DE has to be positive integers.\nBy trial we take CD = 2, DE = 9 [since 2^2 + 9^2 = 4 + 81 = 85]\n\nThus we have the length of four sides of the hexagon ABCDEF,\nAB = 6, BC = 7, CD = 2, DE = 9.\n\nNow, let us concentrate on triangle AFE.\nIn any triangle, the sum of any two sides is always greater than the third side.\nSo, in triangle AFE, AF + FE > AE\ni.e. AF + FE > sqrt(85) = 9.22 [since AE = AC sqrt(85)]\n\nAs length of 4 sides of the hexagon are fixed so for minimum perimeter the sides of triangle AFE has got to be minimum. As AE = sqrt(85) is also fixed so the other two sides has to be minimum and hence their sum has to be minimum i.e. (AF + FE) has to be minimum. But AF + FE > 9.22 and also both AF and FE has to be positive integers, \nso (AF + FE)min = 10.\n\nNow, 10 = 1+9 = 2+8 = 3+7 = 4+6 = 5+5 \n\nSides AF and FE cannot attain any pair of these values (1,9; 2,8; 3,7; 4,6; 5,5) as length of each side of the hexagon has got to be different positive integers and we already have the nos. 2, 6, 7, 9 as length of 4 sides.\n \nSo, AF + FE = 11 [next positive integer to 10]\nAgain 11 = 1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6\n\nThus, AF and FE can attain 2 pairs of values, either 1 and 10 or 3 and 8. \nSo the minimum value of (AF + FE) of possible values of AF and FE is 11.\nHence the minimum value of perimeter of the hexagon is \nAB + BC + CD + DE + EF + (FE + AF)\n= 6 + 7 + 2 + 9 +11\n= 35 \n\nPART (b)\nTo find a minimum value of AC, AC has to be less than present value sqrt(85). Further there must be two pairs of positive integers whose sum equals AC^2 (which is less than 85). So those positive integers has got to be in the range 1 through 9. The squares of these positive integers are presented below. One cannot find two pairs from these numbers such that the sum of each pair is equal to any number less than 85. \n1 4 9 16 25 36 49 64 81\nSo the minimum value of AC is sqrt(85) and the corresponding perimeter of the hexagon is 35.\n\nTo get the figure of this hexagon give me your email id so that I can send it as attachment.
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c = 108 degrees\nyou can find this since A + B + c = 180\nA=B+C+72\nThensubstitute to get A + A + A + 72 = 180\n3A + 72 = 180\nA = 36 so then C= 108
222717
I think that what you really want here is the tangent ratio. You know the slope of the line, right (hint, it is right there in the equation)? You know that the tangent ratio is a over b (where a and b are the non-hypotenuse sides of the triangle), right? You should be able to use that to figure out the tangent of the angle made by the line. Now all you have to do is figure out the angle that has that number as a tangent. That would be the "tan -1" (inverse tangent) function on your calculator.\n\nIf you are working from trig tables rather than a calculator, find the number in the second column of the tangent table that is closest to your tangent number then look in the first column for the angle.\n\nIf you are required to use sin and cosine remember your Pythagorean theorem and use it to find the hypotenuse.\n\nASIDE TO Drakestone: Your solution only finds the sine of the angle, not the angle itself. You need to apply "sin -1" to your answer to get the final answer.\n\n\nASIDE TO eternal...: You are wrong too. graph y=5x+3 and you will see that it is way too steep to be an 11 degree angle. You mixed up your opposite and adjacent values in calculating your tangent.
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If u r saying that u have three sides of a triangle as a,b,c\nAnd u want to determine angle b/w them, u can use the formula\ncosA=(b^2+c^2-a^2)/2bc where A is the angle formed by the sides b and c.similarly if B is the angle formed by a and c, it can be calculated as\ncosB=(a^2+c^2-b^2)/2ac\nonce u have calculated A and B, subtract their sum from 180, u will get third angle.\nRemember u will have to use inverse table for cos to find A and B