How do you use law of sines to solve the triangle given c= 59ft, A= 65 degrees, B= 64 degrees?

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How do you use law of sines to solve the triangle given c= 59ft, A= 65 degrees, B= 64 degrees?

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Answer 1 · 2016-11-24T11:33:24

Solution: $∠ A = 65^{0}, ∠ B = 64^{0}, ∠ C = 51^{0}, a = 68.81 f t, b = 68.24 f t, c = 59 f t$ $/_A=65^0 , /_B=64^0 ,/_C=51^0, a=68.81 ft, b= 68.24 ft,c=59ft$

Explanation:

$∠ C = 180 - (65 + 64) = 51^{0}; c = 59 f t$ $/_C=180-(65+64)=51^0 ;c= 59 ft$

Applying sine law $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin c}$ $a/sinA=b/sinB=c/sinc$ we get

$a = c \cdot \frac{sin A}{sin C} = 59 \cdot \frac{sin 65}{sin 51} = 68.81 (2 d p)$ $a=c*sinA/sinC=59*sin65/sin51=68.81(2dp)$ and
$b = c \cdot \frac{sin B}{sin C} = 59 \cdot \frac{sin 64}{sin 51} = 68.24 (2 d p)$ $b=c*sinB/sinC=59*sin64/sin51=68.24(2dp)$

Solution: $∠ A = 65^{0}, ∠ B = 64^{0}, ∠ C = 51^{0}, a = 68.81 f t, b = 68.24 f t, c = 59 f t$ $/_A=65^0 , /_B=64^0 ,/_C=51^0, a=68.81 ft, b= 68.24 ft,c=59ft$ [Ans]

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How do you use law of sines to solve the triangle given c= 59ft, A= 65 degrees, B= 64 degrees?

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