Question

I need help with these two trigonometry problems.

Answer 1

Answer:

$B \approx 32.1\°; A \approx 39.0\°$

Step-by-step explanation:

Sine theorem: in any triangle, the ratio between a side and the sine of the opposite angle is constant

$\frac{sin\ \alpha}a = \frac{sin\ \beta}b =\frac{sin\ \gamma}c$

In our case, for the left triangle,

$\frac {sin\ 103\°}{11} =\frac{sin\ B} 6 \rightarrow sin\ B = \frac6{11} sin\ 103\°$

Time to grab a calculator and crunch numbers. Double check your calculator is in degrees and not in radians (plug in sin 30°, if you're getting 0.5 you're good) and you will get

$sin\ B \approx 0.53 \rightarrow B \approx 32.1\°$

Same difference with the right triangle. With the same calculations

$sin\ A = \frac{26}{41} sin 83\° \approx 0.68 \rightarrow A \approx 39.0\°$