7. For this triangle you are given two sides and one angle between the two known sides, from the known information you have, you now know you should use the Law of Cosines to find the missing information which in this case is side c, angle A, and angle B. Side a = 12, side b = 14, and angle C = 110°.
c2 = a2 + b2 – 2ab CosC
c2 = 12 + 14 – 2 (12)(14) Cos110°
c2 = 144 + 196 – 2 (168) Cos110°
c2 = 340 – 336 Cos110°
c2 = 340 + 115
c2 = 455
c = 21.3
Now, three sides and an agle opposite one of the sides are known, so you could decide whether to use the Law of Sines or the Law of Cosines, but the Law of Sines is easier to work with.
a = c
SinA SinC
12 = 21.33
SinA Sin110°
21.33 SinA = 12Sin110°
SinA = 12Sin110°
21.33
SinA = 0.5287
A = Sin 0.5287
ÐA = 32°
Now Angle B can again be found from subtracting angle A and angle C from 180.
180° – 110° – 32° = 38°
ÐB = 38°
8. Just to become more familiar, we will show you one more example of how to find the area of a non-triangular shape, being a quadrilateral in this case.
Now that the two triangles have been formed we can proceed to use the Law of Cosines to find the length of side c.
c2 = a2 + b2 – 2ab CosC
c2 = 102 + 182 – 2 (10)(18) Cos85°
c2 = 100 + 324 – 2 (180) Cos85°
c2 = 424 – 360 Cos85°
c2 = 424 – 31.38
c2 = 392.62
c = 19.8
Now again you must find the measurement for the missing angle on the other triangle in order to solve for the area. Again you can use the Law of Cosines.
c2 = a2 + b2 – 2ab CosC
19.82 = 142 + 162 – 2(14)(16) CosC
392.04 = 196 + 256 – 2(224) CosC
392.04 = 452 - 448 CosC
-59.96 = -448 CosC
0.1338 = CosC
Cos-1 0.1338 = C
ÐC = 82°
Now that you know the angles opposite the side separating the two triangles, you can figure out the area of the entire figure.
Area(1) = ½ (ab) SinC
Area = ½ (10)(18) Sin85°
Area = ½ 180 Sin85°
Area = ½ 179.32
Area(1) = 89.662
Area(2) = ½ (ab) SinC
Area = ½ (14)(16) Sin 82°
Area = ½ (224) Sin 82°
Area = ½ 221.82
Area(2) = 110.912
Area(1) + Area(2) = Area(T)
89.66 + 110.91 = 200.572
Total area equals 200.57 units squared.
