Answer
$\sin A = 0.960 \approx 1$
$\cos A = 0.280 \approx 0.300$
$\sin B = 0.280 \approx 0.300$
$\cos B = 0.960 \approx 1$
Work Step by Step
Steps to Answer-
We will use given data about triangle ABC and use Pythagoras Theorem to solve for 'c'-
We know that -
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
$c^{2} = (19.44)^{2} + (5.67)^{2}$
$c^{2} = 377.9136 +32.1489$
$c^{2} = 410.0625$
therefore $ c = \sqrt (410.0625)$ = 20.25
Now we can write the required T-functions of A and B using $a=19.44$ , b = 5.67 and c = 20.25
$\sin A = \frac{a}{c} = \frac{19.44}{20.25}$ = $0.960 \approx 1$
$\cos A = \frac{b}{c} = \frac{5.67}{20.25}$ = $ 0.280 \approx 0.300$
$\sin B = \frac{b}{c} =\frac{5.67}{20.25}$ = $ 0.280 \approx 0.300$
$\cos B =\frac{a}{c} = \frac{19.44}{20.25}$ = $0.960 \approx 1$
