9514 1404 393
Explanation:
It is hard to tell what sort of axioms or theorems you might have available to invoke for this proof. Here, we'll assume that you can use the fact that the sine function of acute angles maintains ordering:
sin(a) > sin(b) iff a > b . . . . . where 0 ≤ a, b ≤ 90°
__
Draw altitude TX, where X lies on RS. The sine relation tells you ...
TX/TS = sin(∠S)
TX/TR = sin(∠R)
Dividing the second equation by the first, we have ...
sin(∠R)/sin(∠S) = (TX/TR)/(TX/TS) = TS/TR
Given that ∠R > ∠S, we know sin(∠R) > sin(∠S), so we have ...
sin(∠R)/sin(∠S) = TS/TR > 1
TS > TR . . . . multiplying by TR
