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Two teams are pulling a heavy chest, located at point x. the teams are 4.6 meters away from each other. team a is 2.4 meters away from the chest, and team b is 3.2 meters away. their ropes are attached at an angle of 110°. triangle a b x is shown. angle a x b is 110 degrees. the length of x b is 3.2, the length of b a is 4.6, and the length of a x is 2.4. law of sines: startfraction sine (uppercase a) over a endfraction = startfraction sine (uppercase b) over b endfraction = startfraction sine (uppercase c) over c endfraction which equation can be used to solve for angle a? startfraction sine (uppercase a) over 2.4 endfraction = startfraction sine (110 degrees) over 4.6 endfraction startfraction sine (uppercase a) over 4.6 endfraction = startfraction sine (110 degrees) over 2.4 endfraction startfraction sine (uppercase a) over 3.2 endfraction = startfraction sine (110 degrees) over 4.6 endfraction startfraction sine (uppercase a) over 4.6 endfraction = startfraction sine (110 degrees) over 3.2 endfraction

Respuesta :

According to law of sine, the part of the equation is sin110/4.6 = sinC/3.2.

How to calculate the value?

According to the law of sines, a/sinA = b/sinB. In this case, it's appropriate to use it for the triangle.

The assigned name is given for each side and angles. Therefore, the part of the equation that the rule is appropriate is sin110/4.6 = sinC/3.2.

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alynj1731

Answer: A

Step-by-step explanation:

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