Answer :
Imagine [tex]\angle A[/tex] at the top of the triangle and [tex]\angle B[/tex] at the bottom. (or used the attached picture for reference)
[tex]\sin A= \frac{opposite}{hypotenuse}[/tex]
[tex]\cos B = \frac{adjacent}{hypotenuse}[/tex]
One thing to note, however, is that the side opposite [tex]\angle A[/tex] is the same side as the one adjacent to [tex]\angle B[/tex]! Thus [tex]\sin A=\cos B[/tex].
Substitute these values for [tex]2x+0.1[/tex] and [tex]4x-0.7[/tex] and solve.
[tex]4x-0.7=2x+0.1 \\ 4x=2x+0.8 \\ 2x=0.8\\ \boxed{x=0.4}[/tex]
[tex]\sin A= \frac{opposite}{hypotenuse}[/tex]
[tex]\cos B = \frac{adjacent}{hypotenuse}[/tex]
One thing to note, however, is that the side opposite [tex]\angle A[/tex] is the same side as the one adjacent to [tex]\angle B[/tex]! Thus [tex]\sin A=\cos B[/tex].
Substitute these values for [tex]2x+0.1[/tex] and [tex]4x-0.7[/tex] and solve.
[tex]4x-0.7=2x+0.1 \\ 4x=2x+0.8 \\ 2x=0.8\\ \boxed{x=0.4}[/tex]
