MR12 Lsn 86 - Forest Hills High School

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Aim: How do we solve problems with both law of sine
and law of cosine?
Do Now: In ABC, mA  46, mB  51 and c = 30cm.
Find to the nearest tenth a) AC b) BC
C
B
A
AC = 23.5
HW: p.580 # 25,26,27,28
BC = 21.7
p.584 # 16,17
To determine the distance between points, A and B,
on opposite sides of a swampy region, a surveyor
chose a point C that was 350 meters from point A
and 400 meters from point B. If the measure of
ACB was found to be 105°, find the distance
across the swampy region, AB, to the nearest meter.
Should we use the Law of Sine or Cosine to solve this problem?
The lengths of two sides and the measure degree of
the included angle are known, therefore, we should
use the Law of Cosine
c  350  400  2  350 400cos105,
2
2
2
c  122500 160000 280000cos105
2
c  282500 72469.3  354969.3,
2
c = 596m
To determine the method we should use
for a particular problem:
Law of Cosine: SAS SSS
Law of Sine:
ASA AAS SSA
Notice that AAA is not good for either law.
In ∆PQR p = 25, q = 12, r = 20. Find each
angle to the nearest degree.
(mP  100, mQ  28, mR  52)
In order to know how much seed to buy for a
triangular plot of land, Kevin needs to know the
area of the plot. The lengths of the sides are 15
feet, 25 feet, and 30 feet.
a. Find to the nearest degree the measure of
the smallest angle
b. Using the answer found in part a), find to the
nearest square foot, the area of the plot of
land.
a. 30°
b. 188
The angle of elevation from a ship at point A to the top
of a lighthouse, point B, is 43. When the ship reaches
point C, 300 meters closer to the lighthouse, the angle
of elevation is 56. Find, to the nearest meter, the
B
height of the lighthouse, BD.
56
43
A
C
300
754 m
D
The beam of a searchlight situated at an
offshore point W sweeps back and forth
between shore points A and B. Point W is
located 12 kilometers from A and 25
kilometers from B. The distance between A
and B is 29 kilometers. Find the measure of
AWB to the nearest tenth degree.
292 = 122 + 252 – 2(12)(25)cos W
600cos W = – 72, W ≈ 96.9
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