1. OK, we want csc A, which is 1/sin A. When we have a right triangle whose hypotenuse is the square root of two times the side, we're where the Pythagoreans were twenty five hundred years ago when they realized the diagonal of the unit square was incommensurate with the side.
In other words we have an isoceles right triangle, 45 degree angles, [tex]a=b=14[/tex] and [tex]c=14 \sqrt{2}[/tex]. So [tex]\sin A = 1/\sqrt{2}[/tex] so [tex]\csc A=\sqrt{2},[/tex] fourth choice.
2. AC adjacent to 23, C is a right angle because 23+67=90. We have hypotenuse AB, so
[tex]\cos A = \dfrac{AC}{AB} \quad \textrm{or} \quad AC=AB \cos A = 7.8 \cos 23[/tex], choice four.
3. The cosine is adjacent over hypotenuse.
[tex]\cos A = 36/60 = 3/5[/tex]
The sine is opposite over hypotenuse,
[tex]\sin B= 36/60 = 3/5[/tex]
First choice
