Law of Sines: A = 5 deg, 40 min, B = 8 deg, 15 min, b = 4.8

[asimon2005]

I need help on these problems because I don't understand this sign for example: 45'.

There are 8 problems I need help on.

In exercise 1-14, use the Law of Sines to solve the triangle.
10. A = 5 degress 40', B = 8 degrees 15', b=4.8

12. C= 85 degrees 20', a = 35, c = 50

14. B = 2 degrees 45', b = 6.2, c = 5.8

28. Height: you are standing 40 meters from the base of a tree that is leaning 8 degrees from vertical away from you. The angle of elevation from your feet to the top of the tree is 20 degrees 50'.

30. Bridge Design: A bridge is to be built across a small lake from a gazebo to a dock. The bearing from the gazebo to the dock is S 41 degrees W. From a tree 100 meters from the gazebo, the bearings to the gazebo and the dock are S 74 Degrees E and S 28 degrees E, respectively. Find the distance from the gazebo to the dock.


In exercises 21-26, find the area of the triangle having the indicated angle and sides.

24. A = 5 degrees 15', b = 4.5, c = 22

26. C= 84 degrees 30', a = 16, b = 20

I need someone check one of my answers.
For this Problem:
In exercise 1-14, use the Law of Sines to solve the triangle.

#6. A = 60 degrees, a=9, c=10

sinC/c=sinA/a
Sin C = c(sinA/a)
Sin C = 10(sin60degrees/9)
C=0.960-
B = 180-60-.96=119.04
b=9/sin(60)(sin 119.04)= approx 9.09

Thanks in advance again.

 

[soroban]

Re: Law of Sines Homework

Hello, asimon2005!

In exercises 21-26, find the area of the triangle having the indicated angle and sides.

\(\displaystyle 24)\;\;A = 5^o15',\;b = 4.5,\;c = 22\)

\(\displaystyle 26)\;\;C= 84^o30',\;a = 16,\;b = 20\)

\(\displaystyle \text{Formula: }\;\text{Area} \;=\;\frac{1}{2}\times\text{(side)} \times \text{(side)} \times \sin\text{(included angle)}\)

\(\displaystyle 26)\;\;\text{Area} \;=\;\frac{1}{2}(16)(20)\sin(84.5^o) \;=\;159.2633917\)