In triangle ABC, angle B = 23 , side c = 43m, and side a = 19m. How do you find side b?

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Answer 1 · 2018-03-29T09:13:10

$b \approx 26.57 m$ $b~~26.57m$ .

By the cosine formula,

$b^{2} = c^{2} + a^{2} - 2 a c \cdot cos B$ $b^2=c^2+a^2-2ac*cosB$ ,

$= 43^{2} + 19^{2} - 2 \cdot 43 \cdot 19 \cdot {cos 23}^{\circ}$ $=43^2+19^2-2*43*19*cos23^@$ ,

$= 1849 + 361 - 1634 \cdot 0.921$ $=1849+361-1634*0.921$ ,

$= 2210 - 1504.105$ $=2210-1504.105$ ,

$= 705.895$ $=705.895$ .

$\Rightarrow b = \sqrt{705.895} \approx 26.57 m$ $rArr b=sqrt705.895~~26.57m$ .