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Proof that the sum of the angles in a triangle is 180 degrees

Proof that the sum of the angles in a triangle is 180 degrees. Theorem If ABC is a triangle then <)ABC + <)BCA + <)CAB = 180 degrees. Proof
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Trigonometry/Proof: Angles sum to 180 - Wikibooks, open books ...

Trigonometry/Proof: Angles sum to 180. From Wikibooks, ... We're going to do exactly the same in proving that the sum of the angles in a triangle is 180 degrees.
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Sum of all inner triangle angles is always 180 - Math Is Fun

Triangles Contain 180 ... This is a proof that the angles in a triangle equal 180°: The top line (that touches the top of the triangle) ...
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Angles in a triangle sum to 180° proof (video) | Khan Academy

Learn the formal proof that shows the measures of interior angles of a triangle sum to 180°. ... Angles in a triangle sum to 180° proof. Triangle exterior angle ...
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How to Prove the Angle Sum Property of a Triangle: 7 Steps

How to Prove the Angle Sum Property of a Triangle. ... To prove that the sum of all angles of a triangle is 180 degrees, ... there is a simple proof that can be ...
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geometry - What's a proof that the angles of a triangle add ...

How would you prove the parallel postulates from the sum of the angles of a triangle being 180 degrees? ... Proof. The sum of the three angles at vertices $A$, ...
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Proof of the angle sum theorem -

Why is the sum of the angles in any triangle equal to 180 degrees? Find an answer to your question here along with a nifty proof of the angle sum theorem. Homepage ...
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Proof: Sum of measures of angles in a triangle are 180 ...

Aug 29, 2013 · Proof that the sum of the measures of the angles in a triangle are 180 Watch the next lesson: Missed the previous ...

Geometry: What is the proof that all the angles in a triangle ...

Is there any proof that the sum of 3 angles of a triangle drawn on a sphere is always found to be ... Why do the angles in a triangle add up to 180 and not 200 or ...
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Proof - Sum of Measures of Angles in a Triangle are 180 - YouTube

Sep 21, 2011 · Proof that the sum of the measures of the angles in a triangle are 180

Suggested Questions And Answer :

find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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Write an indirect proof triangle abc, angle a equal 135 prove that angle b is not equal 45

triangel: 3 angels add up tu 180 deg if 1 angel=135 deg, sum av other 2 gotta be (180-135)=45 deg EVERY angel gotta be > 0 so max size av 1 av em 2 angels gotta be < 45 deg
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Examine the diagram at right. Use the given geometric relationships to solve for x, y, and z. Be sure to justify your work by stating the geometric relationship and applicable theorem

The lst time I did geometry like this was 50 years ago, so I'm a bit vague about the names of geometrical proofs and theorems. I can only give you a hint about what to look for yourself.   You have a triangle with two angles, 23 and 48. Let the third angle be T . 23 + 48 + T = 180   (sum of angles of a triangle add up to 180) T = 109 x = T    (opposite angles ???) x = 109   In the quadrilateral, 81 + z = 180     (because of the two parallel lines) z = 99   Let the angle opposite to y be T2 T2 + x + z + 81 = 360     (sum of internal angles of a quadrilateral add up to 360) T2 + 109 + 99 + 81 = 360 T2 + 289 = 360 T2 = 71 y = T2    (opposite angles   ???) y = 71 The three angles are: x = 109, y = 71, z = 99
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how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
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Proof by the method of contradiction ,if sin a =O then a =k(pi) for any integer k

Assume a≠kπ as the general solution of sin(a)=0. If k is any integer and zero is an integer then k=0 is a possible value for k. But if k=0 then, since a≠kπ on our assumption, a≠0 is a solution for sin(a)=0. In a right-angled triangle sine=opposite side/hypotenuse by definition. So if sin(a)=0 the opposite side length must be permitted to be zero, since no value of the hypotenuse would make the quotient zero. As the length of the opposite side approaches zero, the angle, a, gets smaller. In the limit, the opposite side length becomes zero, and the angle a must also be zero, so a=0 is a solution to sin(a)=0, so contradicting the implication that a≠0. This would eliminate k=0 as an integer. Also, sin(π-a)=sin(a). When sin(a)=0, sin(π)=0. When k=1, if we assume a≠π is a solution we have the contradiction that sin(π)=0. And because sine is a periodic function sin(2π+a)=sin(a) so the argument can be extended to sin(2π), sin(2π+π)=sin(3π)= etc. Therefore we are led to the conclusion that a=kπ is a solution when sin(a)=0.  
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if the sum of the forces in the x-direction = 0 = -0.25w(cos(x)) + w(cos(y)) and the sum of the forces in the y-direction = 0 = 0.25w(sin(x)) + w(sin(y)) - w

From the 1st eqn. -0.25cos(x) + cos(y) = 0 cos(x) = 4cos(y) --------------------- (1) From the 2nd eqn, (1/4)sin(x) + sin(y) - 1 = 0 sin(y) = 1 - (1/4)sin(x) sin^2(y) = (1 - (1/4)sin(x))^2 1 - cos^2(y) = (1 - (1/4)sin(x))^2 Using (1), 1 - (1/16)cos^2(x) = (1 - (1/4)sin(x))^2 1 - (1/16)(1 - sin^2(x)) = (1 - (1/4)sin(x))^2 1 - 1/16 +(1/16)sin^2(x) = 1 - (1/2)sin(x) + (1/16)sin^2(x) 15/16 = 1 - (1/2)sin(x) (1/2)sin(x) = 1/16 sin(x) = 1/8 = 0.125 x = sin^(-1)(0.125) x = 7.181 degrees  
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Geometry Proof Similarity

Triangles PWZ and OWX are similar; PYZ and OXY are similar, because both triangles are right-angled and angles PWZ and OXW are equal; and angles OXY and PYZ are equal (complementary angles). OW/OX=PZ/PW; OY/OX=PZ/PY. OW=OX.PZ/PW and PY=OX.PZ/OY. Therefore, OW/PY=OY/PW=(OP+PY)/(OP+OW) Cross-multiplying: OW.OP+OW^2=OP.PY+PY^2 Rearranging: OP(OW-PY)+OW^2-PY^2=0=OP(OW-PY)+(OW+PY)(OW-PY) Thus: (OW+OP+PY)(OW-PY)=0=WY(OW-PY)=0 Since the diagonal WY is non-zero, OW=PY (QED).    
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A board game requires you to throw a 6 to start. How many throws do you expect to have to make before the throw which delivers you the required 6. ???
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Describe the possible values of x in the figure shown

I am assuming that the figure mentioned is a right-angles triangle. If so, then the hypotenuse is the longest side. i.e. hypotenuse = 2x + 31, since 2x + 31 is greater than either 2x + 1 or x + 16. Using pythagoras' theorem, Hypotenuse squared = sum of squares of other two sides. i.e. (2x + 31)^2 = (2x + 1)^2 + (x + 16)^2 4x^2 + 124x + 961 = 4x^2 + 4x + 1 + x^2 + 32x + 256 x^2 + (4 + 32 - 124)x + (1 + 256 - 961) = 0 x^2 - 88x - 704 = 0 Using the quadratic formula, x = 44 +/- sqrt(165) x = 44 +/-  sqrt(165) = 44 +/-  51.3809 x= 95.38, x = -7.38 If x = -7.38, then one side of the triangle will be 2x + 1 = -14.76 + 1 = negative. We cannot have a negative length, so the negative solution for x should be ignored  
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