Question

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.

Answer 1

Two teams are pulling a heavy chest, located at point X,The teams are 4. 6 meters away from each other , ,The equation can be used to solve for angle A by Law of Sine Here according to the questions the missing angles are ∠A=40.82° and ∠B=29.36°.

Given:-

• Team A is 2. 4 meters away from the chest.
• Team B is 3. 2 meters away from the chest.
• At angle 110° rope is attached.
• A X B = 110 degrees.
• The length of angle X B = 3.
• The length of angle AX = 2.4.

To find the equation can be used to solve for angle A we will check for the given two angles by applying Law of sine,

In the following diagram the values are as follows:-

                    $\rm Angles\;as\;follows\\ \angle A = not \;given\\\angle B = not \;given\\\angle C = 110°\\\\\Sides\; as\; follow\\a = 3.2 unit\\b = 2.4 unit\\c = 4.6 unit$

We will now solve for ∠B,

            $\rm \dfrac{\sin C}{c} = \dfrac{sin B }{b}\\\rm \dfrac{\rm sin 110}{ 4.6} = \dfrac{\rm sin B}{2.4}\\sin B = 2.4\times \dfrac{\sin110}{4.6}\\\sin B = 29.36\textdegree$

Again we will solve for ∠A.

             $\rm \dfrac{\sin C}{c} = \dfrac{sin A }{a}\\\\\rm \dfrac{\rm sin 110}{ 4.6} = \dfrac{\rm sin A}{3.2}\\\\sin A = 3.2 \times \dfrac{\sin110}{4.6}\\\\\sin B = 40.82\textdegree$

Therefore,The equation can be used to solve for angle A by Law of Sine the missing angles are ∠A=40.82° and ∠B=29.36°.

Learn more about Sine Law here : brainly.com/question/4058712