Question

part 6: please assist me with these problems​

Answer 1

Answer:  12) 34°     13) 90°

Step-by-step explanation:

$\text{Law of Sines:}\quad \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

12) Given: a = 10.2,  b = 6.8, A = 122°

$\dfrac{\sin 122^o}{10.2}=\dfrac{\sin B}{6.8}\\\\\\\dfrac{6.8\sin 122^o}{10.2}=\sin B\\\\\\\sin^{-1}\bigg(\dfrac{6.8\sin 122^o}{10.2}\bigg)=B\\\\\\34.4^o=B$

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Law of Cosines: a² = b² + c² - 2bc · cos A

Note: The letters can be swapped

13) Given: a = 3,  b = 4,  c = 5, C = ???

3² = 4² + 5² - 2(4)(5) · cos C

9  = 16 + 25 - 40 cos C

9 = 41 - 40 cos C

-32 = -40 cos C

0.8 = cos C

90° = C