reid’s hot-air balloon is 750.0 m directly above a highway. when reid is looking west, the angle...
DESCRIPTION
75 o x = 2000 m z = m y = 1949 m Y X Z We are looking for Angle Z. We can use Sine Law since we have a Side-Angle pair (Y-y). We can also use both forms of cosine law.TRANSCRIPT
8.5 - Solving Acute Triangle Problems
Reid’s hot-air balloon is 750.0 m directly above a highway.When Reid is looking west, the angle of depression to Exit85 is 75o. Exit 83 is located 2 km to the east of Exit 85.Determine the angle of depression, to the nearest degree,from the balloon to Exit 83.
75o
2 km
Example #1 cont’d75o
2 km75o
75o
750.0 m
x
75o
x = 2000 m
z = 776.46 m
y
𝜃
𝜃
𝜃
𝑦 2=𝑥2+ 𝑧2−2𝑥𝑧 cos𝑌
Y
X
Z
𝑦 2=(2000)2+(776.46)2−2(2000)(776.46 )cos 75𝑜
𝑦 2=4×106+6.03×105−8.04×105
𝑦 2=3.799×106𝑦=1949
Example #1 cont’d
75o
x = 2000 m
z = 776.46 m
y = 1949 m
𝜃Y
X
Z
We are looking for Angle Z.We can use Sine Law since we have a Side-Angle pair (Y-y).
We can also use both forms of cosine law.
Therefore, the angle of depression from the balloon to Exit 83 is about 23o.
Example #2N
The captain of a boat leaves a marina and heads due west for 25 km. Then the captain adjuststhe course of his boat and heads N 30o E for 20 km. How far is the boat from the marina?
We are looking for ‘r’.
𝑟2=(20)2+(25)2−2 (20)(25)cos60𝑜
Therefore, the boat is about 23 km away from the marina.