Use the given information to solve the triangle. If two solutions exists, find both.
A=110 degrees
a=125
b=100
a is greater than b so I think that there is only one triangle. I am not sure where to go from here.
we can find B using the sine law
sinB/100 = sin110/125
sinB = .75175
B = 48.74
then C = 21.26
Now Area = (1/2)(100)(125)sin 21.26 = 2265.99
Thank You!!
To solve the triangle, we can use the Law of Sines and the Law of Cosines. Since we know two angles and one side length, we can use the Law of Sines to find the remaining angles and then use the Law of Cosines to find the remaining side lengths.
1. Law of Sines:
The Law of Sines states that for any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. In other words:
a / sin(A) = b / sin(B) = c / sin(C)
We are given the angle A and the side lengths a and b. Let's start by finding angle B:
sin(B) = (sin(A) * b) / a
sin(B) = (sin(110°) * 100) / 125
sin(B) = 0.819 * 0.8
sin(B) = 0.6552
To find angle B, we take the inverse sine (sin^(-1)) of 0.6552:
B = sin^(-1)(0.6552)
B ≈ 40.2°
Now, we can find angle C, since the sum of the angles in a triangle is always 180°:
C = 180° - A - B
C = 180° - 110° - 40.2°
C ≈ 29.8°
2. Law of Cosines:
The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. In this case, we can use it to find side length c:
c^2 = a^2 + b^2 - 2ab * cos(C)
c^2 = 125^2 + 100^2 - 2 * 125 * 100 * cos(29.8°)
c^2 = 15625 + 10000 - 25000 * 0.882
c^2 ≈ 15625 + 10000 - 22050
c^2 ≈ 35625 - 22050
c^2 ≈ 13575
To find c, we take the square root of both sides:
c ≈ √(13575)
c ≈ 116.53
So, we have found that angle B ≈ 40.2°, angle C ≈ 29.8°, and side length c ≈ 116.53.
Since a is greater than b, there is only one triangle that satisfies the given information.