Post: Missed a maths lesson, help with the work.
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01-16-2011, 03:41 PM
#1
Leo99756
Astro
I have looked on the internet for help with this but couldn't find something with good examples. Does anybody have any good resources for this?
Thanks
.
01-16-2011, 08:13 PM
#2
elfmotat
Rᵤᵥ - ½gᵤᵥR ∝ Tᵤᵥ
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Example: Let's say angle A = 60 degrees and the length of the hypotenuse is 5.
What we need to find the area of the triangle are the lengths of sides a and b.
For side a: We know that sin(A)=a/h.
Plug the values you already know into the equation:
sin(60)=a/5
Now solve for a:
5*sin(60)=a
This means the length of side a is 5*sin(60). Plug this into a calculator:
a=5*sin(60)=4.33
To find side b you can now either use the Pythagorean Theorem, or you can use the cosine function. I'll show you how using cosine:
cos(A)=b/h
cos(60)=b/5
5*cos(60)=b
b=2.5
Finding the area is simple now:
Area=(1/2)*a*b=(0.5)(4.33)(2.5)=5.4125
If you're still stuck, watch these videos:
[ame=https://www.youtube.com/watch?v=F21S9Wpi0y8]YouTube - Basic Trigonometry[/ame]
[ame=https://www.youtube.com/watch?v=QS4r_mqs-rY]YouTube - Basic Trigonometry II[/ame]
I don't want to confuse you with this, but he uses radians in his other videos:
[ame=https://www.youtube.com/watch?v=9zspW8u6kQM]YouTube - Radians and degrees[/ame]
[ame=https://www.youtube.com/watch?v=znR9tW4AiZI]YouTube - Using Trig Functions[/ame]
[ame=https://www.youtube.com/watch?v=RoXmKYjpLGk]YouTube - Using Trig Functions Part II[/ame]
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Example: Let's say angle A = 60 degrees and the length of the hypotenuse is 5.
What we need to find the area of the triangle are the lengths of sides a and b.
For side a: We know that sin(A)=a/h.
Plug the values you already know into the equation:
sin(60)=a/5
Now solve for a:
5*sin(60)=a
This means the length of side a is 5*sin(60). Plug this into a calculator:
a=5*sin(60)=4.33
To find side b you can now either use the Pythagorean Theorem, or you can use the cosine function. I'll show you how using cosine:
cos(A)=b/h
cos(60)=b/5
5*cos(60)=b
b=2.5
Finding the area is simple now:
Area=(1/2)*a*b=(0.5)(4.33)(2.5)=5.4125
If you're still stuck, watch these videos:
[ame=https://www.youtube.com/watch?v=F21S9Wpi0y8]YouTube - Basic Trigonometry[/ame]
[ame=https://www.youtube.com/watch?v=QS4r_mqs-rY]YouTube - Basic Trigonometry II[/ame]
I don't want to confuse you with this, but he uses radians in his other videos:
[ame=https://www.youtube.com/watch?v=9zspW8u6kQM]YouTube - Radians and degrees[/ame]
[ame=https://www.youtube.com/watch?v=znR9tW4AiZI]YouTube - Using Trig Functions[/ame]
[ame=https://www.youtube.com/watch?v=RoXmKYjpLGk]YouTube - Using Trig Functions Part II[/ame]
The following 2 users say thank you to elfmotat for this useful post:
01-22-2011, 12:25 AM
#6
zb4
Banned
Originally posted by Leo99756
I was off sick and missed a maths lesson in which the class found out how to work out and prove the area of a triangle with 1/2 ab Sin C.
I have looked on the internet for help with this but couldn't find something with good examples. Does anybody have any good resources for this?
Thanks.
I have looked on the internet for help with this but couldn't find something with good examples. Does anybody have any good resources for this?
Thanks
.
gay!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
01-22-2011, 04:18 AM
#8
Millz
Worth the Weight
Originally posted by elfmotat
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Example: Let's say angle A = 60 degrees and the length of the hypotenuse is 5.
What we need to find the area of the triangle are the lengths of sides a and b.
For side a: We know that sin(A)=a/h.
Plug the values you already know into the equation:
sin(60)=a/5
Now solve for a:
5*sin(60)=a
This means the length of side a is 5*sin(60). Plug this into a calculator:
a=5*sin(60)=4.33
To find side b you can now either use the Pythagorean Theorem, or you can use the cosine function. I'll show you how using cosine:
cos(A)=b/h
cos(60)=b/5
5*cos(60)=b
b=2.5
Finding the area is simple now:
Area=(1/2)*a*b=(0.5)(4.33)(2.5)=5.4125
If you're still stuck, watch these videos:
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You must login or register to view this content.
I don't want to confuse you with this, but he uses radians in his other videos:
You must login or register to view this content.
You must login or register to view this content.
You must login or register to view this content.
You must login or register to view this content.
You must login or register to view this content.
You must login or register to view this content.
You must login or register to view this content.
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Example: Let's say angle A = 60 degrees and the length of the hypotenuse is 5.
What we need to find the area of the triangle are the lengths of sides a and b.
For side a: We know that sin(A)=a/h.
Plug the values you already know into the equation:
sin(60)=a/5
Now solve for a:
5*sin(60)=a
This means the length of side a is 5*sin(60). Plug this into a calculator:
a=5*sin(60)=4.33
To find side b you can now either use the Pythagorean Theorem, or you can use the cosine function. I'll show you how using cosine:
cos(A)=b/h
cos(60)=b/5
5*cos(60)=b
b=2.5
Finding the area is simple now:
Area=(1/2)*a*b=(0.5)(4.33)(2.5)=5.4125
If you're still stuck, watch these videos:
You must login or register to view this content.
You must login or register to view this content.
I don't want to confuse you with this, but he uses radians in his other videos:
You must login or register to view this content.
You must login or register to view this content.
You must login or register to view this content.
Would it be possible to explain that in English?
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