The hypotenuse, c, of right $\triangle$ABC is $7.0$cm long. A trigonometric ratio for angle $A$ is given for four different triangles. Which of these triangles has the greatest area?
a) sec $A$ = $1.7105$
b) cos $A$ = $0.7512$
c) csc $A$ = $2.2703$
d) sin $A$ = $0.1515$
Hint: Imagine a semicircle with diameter $AC = 7$. Out of all the possible right-angled triangles that fit in the semicircle (where the right angle $B$ is somewhere on the circumference), the one with largest possible area is when $ABC$ is isosceles so that $A = 45^\circ$. So take each trig ratio and use trig identities to "standardize" them so that they are all, say, in terms of $\cos A$. Then find the one which is closest to: $$ \cos 45^\circ = \frac{\sqrt 2}{2} = 0.7071 \ldots $$