Question

Any help someone.... (solve triangles using the law of sines)

Answer 1

Answer:

$B=51$°

Step-by-step explanation:

Use the Law of Sines as follows:

$\frac{sinC}{c} =\frac{sinB}{b}$

Insert the appropriate values:

$\frac{sin100}{14} =\frac{sinB}{11}$

Isolate B. Multiply both sides by 11:

$11*(\frac{sin100}{14} )=11*(\frac{sinB}{11})\\\\\frac{11*sin100}{14} =sinB\\\\sinB=\frac{11*sin100}{14}$

Use the inverse:

$B=sin^{-1}(\frac{11*sin100}{14})$

Insert the equation into a calculator and round to the nearest degree:

$B=50.7\\\\B=51$

Angle B is 51°.

:Done