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the side of the triangle is equal to 2
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and the two adjacent angles are 30° and
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45° find the other two sides of the
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triangle as always before watching
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further try to solve this problem on
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your own from the vertex of the triangle
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drop a perpendicular to its side which
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according to the conditions is equal to
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two that is draw the height of the
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triangle in this way this height will
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divide the given triangle into into two
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right triangles let the height of the
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X then considering that in a right
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triangle the side opposite the 30° angle
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is equal to half of the hypotenuse the
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hypotenuse will be equal to 2x now let's
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move on to the second right triangle
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where one of the acute angles is equal
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45° this means that the other acute
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45° thus the right triangle turns out to
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be isoceles which means the second leg
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will also be equal to X but then to find
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the leg of the other right triangle we
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need to subtract X from the entire side
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of the original triangle which is equal
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to 2 so the leg of the other right
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triangle will be equal to 2 - x
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furthermore we can immediately Express
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the side of the second right triangle
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that we need to find considering that it
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is the hypotenuse of the second right
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Tri triangle whose legs are equal to X
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this hypotenuse can be found using the
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Pythagorean theorem suppose we denote it
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as y then the square of the hypotenuse
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will be equal to the sum of the squares
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of the legs simplifying the terms and
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taking the square root we get the
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hypotenuse in the form of x * the < TK
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of 2 so the second side of the original
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triangle we need will be equal to x *
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theare < TK of 2 let's write this down
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next according to the Pythagorean
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theorem for the first right triangle we
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x² the square of the hypotenuse that is
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2x^2 will be equal to the sum of the
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squares of the legs expand the
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parentheses move everything to the left
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side and combine like terms after
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dividing both sides of the equation by
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two we get a quadratic equation in the
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form of x^2 + 2x - 2 = 0 we solve this
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using the discriminant or we can solve
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it using D1 which is the so-called
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quarter discriminant D1 which is equal
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to the discriminant ID 4 will be equal
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to B / 2^ 2 which is 1 minus a which is
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+ 2 we get three the roots of such a
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quadratic equation will be equal to
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minus B / 2 which is -1 plus orus the
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square < TK of D1 which is plus or minus
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the squ < TK of 3 considering that X
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which represents the length of the
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segment must be positive we leave only
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one possible root so X will be equal to
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the < TK of 3 - 1 now we can find the
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necessary sides of the original triangle
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one of its sides was equal to
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2x multiply 2 by the < TK of 3 - 1 and
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get 2 square roots of 3 - 2 the second
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side was equal to x * < TK 2 multiply
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the < TK of 2 by the < TK of 3 - 1 and
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get the < TK 6 - the < TK 2 so the sides
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of the triangle are two square roots of
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3 - 2 and the square < TK of 6 - theare
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< TK of 2 the problem is solved if you
3:46
understood the solution give a thumbs up
3:49
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