In Exercises 35–36, the three given points are the vertices of a ...

Question

In Exercises 35–36, the three given points are the vertices of a triangle. Solve each triangle, rounding lengths of sides to the nearest tenth and angle measures to the nearest degree.

A(0, 0), B(-3, 4), C(3, -1)

Accepted Answer

Given the vertices of a triangle, determine the lengths of all its sides and measures of all its angles round the side links and the angles to one decimal place where we have three points negative 20, negative 65 and four negative two. No, we can make a trip out of these on a coordinate plane. If I were to make this cord a plane just as a rough little sketch, you have a point at negative 65 a point and 820. And the point at four negative two. What an order that is point A point B and point C. Let's first determine the side links. We can do this by using the distance formula. The distance formula is given by the square root of X two minus X one squared plus Y two minus Y one squared. We can find a distance between every one of these points. Let's say we go A B first that will be the square root of negative six minus negative two squared plus five minus zero squared. We will then get the square root four squared plus five square which gives us the square root of 41. Let's check the distance of BC. Now, that would be the square root of four minus negative six squared plus negative two minus five squared. That is the square root of 10 squid plus seven squared, which is the square root of 1 49. And finally, we will find AC, which is the square root A four minus negative two squared plus, take the two minus zero square. That gives the square root a six square plus two square, which gives us the square root of 40. We can round all of this to get 6. 12.2 and 6.3. Now those are over side links which I will redraw our triangle over here to 6.3 and 6.4. Now we need to find our angles. We can first use this law of cosines to find one of our angles. This equation is given by or angle A A square equals B squared plus C squared minus two BC oh sign of eight. Now we're denoting our first side side A as our longest side. So we will say 12.2 squared equals are sides B and C which we will also denote this side is B. This side is C 6.3 squared plus 6.4 squared minus two, multiplied by 6.3 6.4 all multiplied by cosine A. Now we can simplify all of this. At the end, we should get a equals 1 47. degrees. If we were to simplify all of this and solve for a, we cannot make use of a lot of signs to get the rest of our angles changing our triangle. Again, we will make angle A, the angle directly opposite of our longest side. A angle B will be directly opposite of side B and angle C will be opposite side C. Now, we can use a lot of signs to find our last two angles. A lot of science tells us we can take a divided by a sign of angle A and instead it equals to B divided by S sign of angle B. And also can set that equals to C divided by a sine of angle C. Let's look for angle B first, we have side A which is 12.2 divided by a sign of angle A sine of 1 47.1 degrees. No, we have side beep which is 6.3 divided by sign can go B we now need to solve for B, we would get sign B is equals to 6.3 sign 47.1 all divided by 12.2. Now to get B, we will take the inverse sun or the arc sign which we can write as B equals sine to the negative. First of 6.3 sign 47.1 divided by 12.2. We then get B as equals to 16. degrees. Finally, we can find angle C by doing the same thing. And we actually don't have to use a lot of signs. Again, if we know that the sum of the angles in the triangle equals 1 80 we can also say 1 80 is equals to 1 47. plus 16.3 plus C giving AC is equals to 16.6 degrees. We now have everything for our answer. We have side A, it's 12.2. Side B is 6.3. Side C is 6.4. Angle A is 1 47.1 angle B which you went a lower case for the side. By the way, had to change that the angle B will be 16.3 and angle sea is 16.6 degrees. We have all of our side links and all of our angles for this problem. OK? I hope to help you solve the problem. Thank you for watching. Goodbye.

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In Exercises 35–36, the three given points are the vertices of a triangle. Solve each triangle, rounding lengths of sides to the nearest tenth and angle measures to the nearest degree. A(0, 0), B(-3, 4), C(3, -1)