Trigonometric Law of Sines Today’s lessons, we’ll cover the law of sines. The objectives for today is use the law of sines to solve triangles and then find the area of any triangle given the lengths of two sides and the measure of the included angle. Here, we’re going to try to solve oblique triangles. We’ve done earlier to solving all the sides and angles for a right triangle, now, we’re going to cover for oblique triangles. Here, we have a triangle and another triangle. What you should see in common is that the angles are the same but you can’t solve the triangle since there are many solutions to this. So, if you’re given any three angles like here, here or here, you can’t find a unique triangle associated with those three angles. So in other words, knowing the three angle does not tell you about the size. Let’s say we’re given at this triangle a, b, and c and its associated sides, little a, little b, little c, and we’re going to just drive the law of sines in any triangle A, B, C shown here. Let’s say we draw an H that is perpendicular from A to its associated opposite sides such that it’s perpendicular. We’ll call that H, the height of this triangle. What we did here is that we formed two right triangles and therefore we can start using the definitions of the basic trick functions. So in this case, we formed this triangle right here. On the left side of H and on the right side, we formed triangle A, B and where it intersects here. So now, what we have is the sin(C). This angle C is the opposite over the hypotenuse so sin(C)=h/b. Based on this, h must equal to multiplying b on both sides of this equation. The b’s cancels out on the right side of this expression and therefore we have what’s left as b*sin(C). Also, we note for the other triangle that the sin(B) is the opposite over the hypotense. In this case, it’s h/c. Similarly, we can say that now h is equal to C*sin(B). We note that these two expressions are equal so we can set this expression and this expression equal to each other as shown here. Now, dividing respectively by the sin of C on both sides and the sin(B) on both sides, we have the following relationship, that is we have all the variables associated with the angle C and its opposite side C is equal to the ratio of b/sin(B). This is known as the Law of Sines as we’ll see on the next slide where we have the Law of Sines for all three sides namely that a/sin(A)=b/sin(B)=c/sin(C). This is the Law of Sines for any triangle a, b, c. So therefore, you can also take the reciprocal of each of there terms so you could have sin(A)/a, sin(B)/b and sin(C)/c. Let’s try to solve a triangle using the Law of Sines as an example. So here, we have a triangle with angle B equal to 38 degrees and angle C equals 21 degrees. And we note that little b is also 24. This is what’s given. So, whenever you see a side B and angle cup B for example we can trigger that to give us insight that we should use that Law of Sines to solve the triangle. So, here is a picture of what we’re given. And in this case, we know that 38 degrees and 21 degrees plus some angle A is equal to 180 and we can solve for that. But using the Law of Sines as shown here, we can solve for that triangle now. So, let’s start of with what we are given where b is given as 24 and sin(B)=sin(38). With that, we can solve for little c since the angle C is given as 21 degrees. After performing this calculation right here, you get the final answer of little c to b 14. We can follow the same process with the other sides for example a. Now a as I said before can be calculated using the 180-degree rule. So, 180-38-21 yields 121. Now, you can substitute this 121 into this expression here. It says Law of Sines and we can solve for a and we’re given that a=33. that’s what we solved. So therefore, we solved the entire triangle, found all the angles that’s what we need so we found angle A, we found little a and we found little c. So here, we can show you the calculation where we solve