I need help with these kind of problems. We are dealing with word problems and/or pictures that are solved by the Law of Sin or Cosine. If any one kows a wonderful web site that can help with these, I would greatly appreciate it.
A parollelogram has side lengths 3 and 5, and one angle is 50 degrees. Find the lengths of the diagonals.
Shorter diagonal: the angle between sides 3 and 5 is 50º (SD for short diagonal)
SD^2 = 5^2 + 3^2 - 2(5)(3)cos 50º
take it from there.
Let me know what you get for the equation of the longer diagonal.
I got 3.8, we round to the nearest tenths.
3.8 was my answer for the short diagonal.
I'm not sure how to find the long diagonal.
for the longer, same sides except the angle between them is now 130º
LD^2 = 5^2 + 3^2 - 2(3)(5)cos130
= 25 + 9 - 30(-.642788)
= 7.2996
= 7.3
Notice this time the last term was actually added because of the double negatives.
last two lines should have been
LD = 7.2996
= 7.3
(I still had it equal to LD^2)
To solve this problem using the Law of Sines or Law of Cosines, we need to find the lengths of the diagonals of the parallelogram. However, the Law of Sines and Law of Cosines are used specifically for solving problems involving triangles.
In this case, you can use the properties of a parallelogram to help solve the problem. A parallelogram has opposite sides that are equal in length and opposite angles that are equal in measure. We will utilize these properties to find the lengths of the diagonals.
Given that the parallelogram has side lengths of 3 and 5, and one angle is 50 degrees, we can use the properties of a parallelogram to find the lengths of the diagonals.
First, let's consider the diagonals of the parallelogram. Let's label one diagonal as AC and the other diagonal as BD.
From the given information, we know that the opposite sides of the parallelogram are equal in length. Therefore, side AC is equal to side BD. In this case, side AC has length 3, and side BD has length 5.
Now, let's consider the diagonals of the parallelogram individually.
For diagonal AC, we can use the Law of Cosines to find its length. The Law of Cosines states:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, a = 3, b = 5, and C is the angle opposite side AC in the triangle formed by the diagonals. Since we do not know the measure of angle C directly, we can use the fact that opposite angles in a parallelogram are equal to find it.
Given that one angle of the parallelogram is 50 degrees, we know that the opposite angle is also 50 degrees. Therefore, angle C is also 50 degrees.
Using the Law of Cosines, we can calculate the length of diagonal AC:
AC^2 = 3^2 + 5^2 - 2 * 3 * 5 * cos(50)
AC^2 = 9 + 25 - 30cos(50)
AC^2 = 34 - 30cos(50)
Next, let's consider diagonal BD. Since opposite sides of a parallelogram are equal, the length of diagonal BD is also 5.
Therefore, the lengths of the diagonals AC and BD are given by:
AC = sqrt(34 - 30cos(50))
BD = 5
To summarize, the length of diagonal AC is sqrt(34 - 30cos(50)), and the length of diagonal BD is 5.
Now, to find a good website that can help with these types of problems, I recommend searching for online math resources such as Khan Academy, Mathway, or Mathisfun. These websites offer explanations, examples, and practice problems for a variety of math topics including trigonometry, which includes the Law of Sines and Law of Cosines.