The line from the point to its reflection should be perpendicular.If we imagine a line from (5,7) to (2,2), it would have a slope of (2-7)/(2-5) = 5/3.
For that line to be perpendicular to y=-2/5x+6, their slopes should be each other's negative reciprocals.
-2/5 negative reciprocal is 5/2, which is not equal to our calculated 5/3, so (2,2) cannot be the reflected point. Evan was wrong.
Q1-74: AB have slope 6/5, C has -5/6, D 6/5, E -5/6 again.
The answer would be 660.4 millimeters
We have the formula tan( a- b ) = ( tan a - tan b ) / ( 1 + tan a· tan b );
We use the formula (sin x)^2 + (cos x)^2 = 1;
Because sin a = 4/5 we have cos a = 3/5; then tan a = sin a / cos a = 4 / 3;
Because sin b = 5 / 13 we have cos b = 12 / 13; then tan b = sin b / cos b = 5 /12;
Finally, tan ( a - b ) = ( 4 / 3 - 5 / 12 ) / ( 1 + (4 / 3)·( 5 / 12 ) ) = ( 9 / 12 ) / ( 1 + 5 / 9 ) = ( 3 / 4 ) / ( 14 / 9 ) = ( 3 / 4 )·( 9 / 14 ) = 27 / 56.
