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Applying Sine to triangles.
- Thread starter Cake
- Start date
D
Deleted member 4993
Guest
Hey Guys,
For number 15)
I found out what a is by 35 = a / 12
I found out the sine value of it which is .5736 then I multiply by 12 to find a = 6.8832
Can someone tell me how to find B?
Same way you found 'a', except
Use
cosine(36°) = b/12
Same way you found 'a', except
Use
cosine(36°) = b/12
Hey Khan,
Sweet, thanks. Do you know the reason why I have to use cosine?
D
Deleted member 4993
Guest
Hey Khan,
Sweet, thanks. Do you know the reason why I have to use cosine?
Well you don't have to!
You know the length of one leg of the triangle (a), you know the length of the hypotenuese(12), you can call Pythagorus to solve for the other leg (b).
HallsofIvy
Elite Member
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- Jan 27, 2012
- Messages
- 7,763
Cake, when you first learned "sine" and "cosine" you should have learned that sine is "opposite side over hypotenuse" and cosine is "near side over hypotenuse". In problem one, you are told the length of the hypotenuse, 12 in, angle 35 degrees, and are asked to find (a), the side opposite the given angle: "opposite side over hypotenuse" is a/12= sin(35). You are also asked to find (b), the side near the angle: "near side over hypotenuse" is b/12= cos(35).
But, as Subhotosh Khan says, you do NOT "have to" use that. Once you know (a) you can use the Pythagorean theorem: \(\displaystyle a^2+ b^2= c^2\) where c= 12 and you have already found a: \(\displaystyle b^2= 12^2- a^2\), \(\displaystyle b= \sqrt{144- a^2}\).
But, as Subhotosh Khan says, you do NOT "have to" use that. Once you know (a) you can use the Pythagorean theorem: \(\displaystyle a^2+ b^2= c^2\) where c= 12 and you have already found a: \(\displaystyle b^2= 12^2- a^2\), \(\displaystyle b= \sqrt{144- a^2}\).