### Video Transcript

In the figure, a horizontal force of magnitude 890 newtons is acting on a particle at πΆ, which is attached to two strings connected to π΄ and π΅, respectively. Given that the particle is in equilibrium and the two strings and the particle all lie in the same vertical plane, find the tension in the two strings to the nearest newton.

In this question, we have three forces acting at point πΆ. We have a force of magnitude 890 newtons acting horizontally to the right together with two tension forces π sub one and π sub two. One way of calculating the values of π sub one and π sub two is using Lamiβs theorem. This states that when three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. If the three forces are π΄, π΅, and πΆ, then π΄ over sin πΌ is equal to π΅ over sin π½, which is equal to πΆ over sin πΎ, where πΌ is the angle between the forces π΅ and πΆ, π½ is the angle between forces π΄ and πΆ, and πΎ is the angle between the forces π΄ and π΅.

In our question, we can begin by calculating the missing angle in the triangle. We know that angles in a triangle sum to 180 degrees. 45 plus 35 is equal to 80 and subtracting this from 180 gives us 100. The missing angle in the triangle is equal to 100 degrees. As the 890-newton force acts horizontally, we have two right angles, as shown in the diagram. Using our knowledge of alternate angles, the angle between the vertical line and the tension force π sub two is 35 degrees. And the angle between the vertical line and the tension force π sub one is 45 degrees. This means that the angle between the 890-newton force and π sub one is 135 degrees. And the angle between the 890-newton force and π sub two is 125 degrees.

We now have the values of the three angles πΌ, π½, and πΎ at point πΆ. Substituting all of our values into Lamiβs theorem gives us 890 over sin 100 degrees is equal to π sub one over sin 125 degrees, which is equal to π sub two over sin of 135 degrees. We begin by focusing on the first two expressions. Multiplying through by sin of 125 degrees, we have π sub one is equal to 890 over sin of 100 degrees multiplied by sin of 125 degrees. This is equal to 740.292 and so on. We are asked to give our answer to the nearest newton. π sub one is therefore equal to 740 degrees.

We can repeat this process to calculate the value of π sub two. This time, we begin by multiplying through by the sin of 135 degrees. This gives us π sub two is equal to 639.033 and so on. Once again, we round to the nearest newton. π sub two is equal to 639 newtons.

The tension in the two strings connected to π΄ and π΅ are 740 newtons and 639 newtons, respectively.