Question

Two squares and an equilateral triangle,find the angle​

Answer 1

☑️We draw the perpendicular MM' to CB and we consider the right triangles MCM' and MBM' and the tangent of angles MCM' and MBM'.

We draw the perpendicular MM' to CB and we consider the right triangles MCM' and MBM' and the tangent of angles MCM' and MBM'. tan(MCM') = MM' / CM'

We draw the perpendicular MM' to CB and we consider the right triangles MCM' and MBM' and the tangent of angles MCM' and MBM'. tan(MCM') = MM' / CM' tan(MBM') = MM' / BM'

☑️The above may be written as

The above may be written as CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM')

The above may be written as CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM') Note that the size of angle MBM' is equal to 75 degrees. Let the length of the side of the square be x.

MM' = x / 2 and CM' + BM' = x

MM' = x / 2 and CM' + BM' = x Substitute the above in the equations CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM')

MM' = x / 2 and CM' + BM' = x Substitute the above in the equations CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM') x = (x/2) / tan(MCM') + (x/2) / tan(75 degrees)

MM' = x / 2 and CM' + BM' = x Substitute the above in the equations CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM') x = (x/2) / tan(MCM') + (x/2) / tan(75 degrees) Divide all terms of the equation by x and multiply them by 2 to obtain

MM' = x / 2 and CM' + BM' = x Substitute the above in the equations CM' = MM' / tan(MCM') and BM' = MM' / tan(MBM') x = (x/2) / tan(MCM') + (x/2) / tan(75 degrees) Divide all terms of the equation by x and multiply them by 2 to obtain 2 = 1 / tan(MCM') + 1 / tan(75 degrees

tan(75 degrees) is calculated using the tangent of a sum formula follows

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) =

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3)

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3) Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3) Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain 1 / tan(MCM') = 2 - 1 / [ 2 + sqrt(3) ]

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3) Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain 1 / tan(MCM') = 2 - 1 / [ 2 + sqrt(3) ] Simplify the above and solve for tan(MCM') to obtain

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3) Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain 1 / tan(MCM') = 2 - 1 / [ 2 + sqrt(3) ] Simplify the above and solve for tan(MCM') to obtain tan(MCM') = 1 / sqrt(3)

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3) Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain 1 / tan(MCM') = 2 - 1 / [ 2 + sqrt(3) ] Simplify the above and solve for tan(MCM') to obtain tan(MCM') = 1 / sqrt(3) Solve the above trigonometric eqution to find the size of angle MCM'

tan(75 degrees) is calculated using the tangent of a sum formula follows tan(75 degrees) = tan(30 degrees + 45 degrees) = [ tan(30) + tan(45) ] / [ 1 - tan(30)*(tan(45) ] = 2 + sqrt(3) Substitute tan(75 degrees) by 2 + sqrt(3) into the equation 2 = 1 / tan(MCM') + 1 / tan(75 degrees) to obtain 1 / tan(MCM') = 2 - 1 / [ 2 + sqrt(3) ] Simplify the above and solve for tan(MCM') to obtain tan(MCM') = 1 / sqrt(3) Solve the above trigonometric eqution to find the size of angle MCM' MCM' = 30 degrees

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Answer 2

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