Law of sine and cosine

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Ihatdogs10

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Greetings can someone help me to solve it?
A and B are two points on opposite sides of a body of water, and soundings are to be taken in the line AB at points one quarter, one half, and three quarters of the distance from A to B. On the shore, a line AC 1200 feet long is measured, and angles BAC = 63° 19' and ACB = 78° 43'. What angles must be turned off from CA at C in order to line up the boat from which the soundings are made at the proper points on AB?
 
Ihatdogs10 said:
Hi I have a question what law should I use to find those angle? Law of cos?
Click to expand...
Have you tried either one?

First you'll need to calculate the length of AB.

In triangle ABC, you know the length of AC and you know two angles. Which law should you use to calculate length of AB?

Please share your work.
 
Last edited by a moderator:
greetings to everyone. i'd like someone to help me show the solution to this question. i am gonna be grateful with all my heart.

A and B are two points on opposite sides of a body of water, and soundings are to be taken in the line AB at points one quarter, one half, and three quarters of the distance from A to B. On the shore, a line AC = 1200 feet long is measured, and angles BAC = 63° 19' and ACB = 78° 43'. What angles must be turned off from CA at C in order to line up the boat from which the soundings are made at the proper points on AB?
 
Subhotosh Khan said:
Have you tried either one?

First you'll need to calculate the length of AB.

In triangle ABC, you know the length of AC and you know two angles. Which law should you use to calculate length of AB?

Please share your work.
Click to expand...
 

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CORRECT (method - I did not check arithmetic).

- now continue on to solving for the smaller angles
 
great job,

now the first angle has this length [MATH]\frac{1 \cdot 1912.87}{4}[/MATH]
second angle has this length [MATH]\frac{2 \cdot 1912.87}{4}[/MATH]
third angle has this length [MATH]\frac{3 \cdot 1912.87}{4}[/MATH]

60.png
 
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nasi112 said:
great job,

now the first angle has this length [MATH]\frac{1 \cdot 1912.87}{4}[/MATH]
second angle has this length [MATH]\frac{2 \cdot 1912.87}{4}[/MATH]
third angle has this length [MATH]\frac{3 \cdot 1912.87}{4}[/MATH]
Click to expand...
Can I ask the formula? Where does the 4 come from? I am so sorry I am lots
 
nasi112 said:
great job,

now the first angle has this length [MATH]\frac{1 \cdot 1912.87}{4}[/MATH]
second angle has this length [MATH]\frac{2 \cdot 1912.87}{4}[/MATH]
third angle has this length [MATH]\frac{3 \cdot 1912.87}{4}[/MATH]
Click to expand...
@nasi112 - Please use "*" to indicate multiplication - instead of "\(\displaystyle \cdot \)". That will avoid confusion with decimal points (.) or period.
 
Ihatdogs10 said:
Can I ask the formula? Where does the 4 come from? I am so sorry I am lots
Click to expand...
divide by [MATH]4[/MATH] means I want quarter the line of AB

you can do it like this also
first angle has a length of [MATH]0.25[/MATH] x [MATH]1912.87[/MATH]
second angle has a length of [MATH]0.5[/MATH] x [MATH]1912.87[/MATH]
third angle has a length of [MATH]0.75[/MATH] x [MATH]1912.87[/MATH]
 
Oh I see I will solve it tom thank you
nasi112 said:
divide by [MATH]4[/MATH] means I want quarter the line of AB

you can do it like this also
first angle has a length of [MATH]0.25[/MATH] x [MATH]1912.87[/MATH]
second angle has a length of [MATH]0.5[/MATH] x [MATH]1912.87[/MATH]
third angle has a length of [MATH]0.75[/MATH] x [MATH]1912.87[/MATH]
Click to expand...
 
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