In the given figure, the force 𝐑 with magnitude 12 newtons is resolved into two components, 𝐅 sub one and 𝐅 sub two. And the force 𝐑 bisects the angle between the directions of 𝐅 sub one and 𝐅 sub two, where the angle between the directions of 𝐅 sub one and 𝐅 sub two is 70 degrees. Find the magnitude of 𝐅 sub two to the nearest newton.
One way of answering this question is to redraw our diagram using our knowledge of the triangle of forces, where the lengths of each side of our triangle are equal to the magnitude of the force 𝐑 together with the components 𝐅 sub one and 𝐅 sub two. We know that the angle 𝑥 between the force 𝐑 and 𝐅 sub one is equal to a half of 70 degrees. This is because force 𝐑 bisects the angle between the directions of 𝐅 sub one and 𝐅 sub two. 70 degrees divided by two is 35 degrees. Therefore, angle 𝑥 equals 35 degrees. This is also the angle between force 𝐑 and 𝐅 sub two.
This means that we have an isosceles triangle. And since angles in a triangle sum to 180 degrees, the third angle is equal to 110 degrees. The law of sines or sine rule states that 𝑎 over sin 𝐴 is equal to 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶, where lowercase 𝑎, 𝑏, and 𝑐 are the side lengths of the triangle and capital 𝐴, 𝐵, and 𝐶 are the angles opposite the corresponding sides.
In this question, we are trying to find the magnitude of 𝐅 sub two. And we know that the magnitude of 𝐑 is 12 newtons. 𝐅 sub two is opposite a 35-degree angle in our triangle. And force 𝐑 is opposite 110 degrees. Substituting these values into the law of sines, we have 𝐅 sub two over sin of 35 degrees is equal to 12 over sin of 110 degrees. We can then multiply both sides of our equation by the sin of 35 degrees. Typing the right-hand side into our calculator gives us 7.324 and so on. We are asked to give our answer to the nearest newton. Rounding to the nearest number gives us an answer of seven. Therefore, the magnitude of 𝐅 sub two to the nearest newton is seven newtons.
Since our triangle is isosceles, it is worth noting that the magnitude of 𝐅 sub one is equal to the magnitude of 𝐅 sub two. So, this is also equal to seven newtons.