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Mars2501 [29]
1 year ago
14

I need help with these two trigonometry problems.

Mathematics
1 answer:
Sloan [31]1 year ago
5 0

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\°

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