Question

In triangle ABC, a=2 ,b=3 ,c=4; find cos B​

Answer 1

Answer:

$cos(B) = \frac{11}{16}$

Step-by-step explanation:

By law of cosines,

${b}^{2} = {a}^{2} + {c}^{2} - 2ac \cos(B) \\ \\ {3}^{2} = {2}^{2} + {4}^{2} - 2 \times 2 \times 4 \times cos(B)$

$16 \: cos(B) = 4 + 16 - 9 \\ \\ 16 \: cos(B) = 11$

$cos(B) = \frac{11}{16}$

Hope it helps