a. Apparently, we're to assume that the fence is as long as the perimeter of the lot. Thus, we must find the length of the 3rd side.
We are given the lengths of two sides and the angle between them, so the Law of Cosines can be used to advantage. If we call the given sides "a" and "b" and the angle between them "C", that relation tells us
... c² = a² + b² - 2ab·cos(C)
Filling in the given dimensions gives
... c² = 100² + 150² - 2·100·150·cos(38°)
... c² = 10,000 + 22,500 - 30,000·0.7880108 = 8859.677
... c = √8859.677 ≈ 94.13
Then the total length of fencing needed is the perimeter (p) of the yard:
... p = a + b + c = 100 m + 150 m + 94.13 m = 344.13 m
b. Given two sides and the angle between them, the area of the triangle can be found from
... A = (1/2)ab·sin(C)
Filling in the given information, you have
... A = (1/2)·(100 m)·(150 m)·sin(38°) = 4617.46 m²
This area is larger than 1 acre (≈ 4047 m²), so they do have enough land for a pool.
c. The largest angle in the yard will be opposite the longest side of the yard. It can be found using the law of cosines.
... b² = a² + c² - 2ac·cos(B)
... cos(B) = (a² +c² -b²)/(2ac) = (100² +94.13² -150²)/(2·100·94.13) ≈ -0.193375
... B = arccos(-0.193375) ≈ 101.15° . . . . largest angle in the yard
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Using the law of cosines avoids the ambiguity that comes with using the law of sines.
