Homework Help

[Starlight Fan 538]

Actually I think I can read it. For the last problem, it is asking for a height of 1.8 meters OR LESS, so I think you want 114, not 6

Oh dear god I didn't read the question properly. Thank god I'm not in an exam XD

 

You've been a great help so far, thank you.

[Starlight Fan 538]

Oh god, not me again...

 

Anyway the question this time is that a straight path is to be built from B to the nearest point on the road AC. I have to find the length of the straight path. Here's the diagram. I calculated angle ABC using the sine rule in a previous part of the question:

 

Edited March 9, 2015 by Sonata Dusk

[IceBeerG 55]

Oh oh, i love mathematic tasks. Back in school i was really good with thous, but now have to look after formula again but... still like to solve them ^^ 

 

First - i doubt that ABC = 54. Can i look at your calculations about this one? (just curious ><)

 

Try this one - S of ABC = 1/2 AC*h. However S of ABC also = 1/2*BC*AC sin BCA. It leads to: 1/2 135*h=1/2*BC*135*~0,44 => h=BC*~0,44, The problem is to fin BC , that can be easily found with angle BAC (in your case is 100) but, again i doubt it is right. If it is, then => BC/sin100=79/sin29=>BC= 79*~0.98/~0,48=>161.3*~0,44 h=>~70,97 (However i find quite strange that there is no absolute number. It's a bit unfamiliar to operate for me. May i ask you what your age, or, better, what is your grade? collage 1st,2nd,3rd? uni? school one of last classes?)

 

I tried to calculate BC somehow negating the fact that ABC is 54, but found this task either to hard to solve for a break or impossible with given information.

Edited March 9, 2015 by IceBeerG

[Astral Jackpot 57]

Oh god, not me again...

 

Anyway the question this time is that a straight path is to be built from B to the nearest point on the road AC. I have to find the length of the straight path. Here's the diagram. I calculated angle ABC using the sine rule in a previous part of the question:

 

IMG_20150309_194138_1425901329681.jpg

 

From what I can gather to figure out that question we would need to know either the angles formed when the new line hits the line AC, where it hits the line AC so as to figure out the length of the two sides it makes or the exact split of the angle it makes at B so as to actually know what angles are formed. If we had either of those 3 it could be done easily.

Also angle B is 48.51 I think

[Night Wing 387]

screw homework, blast those wubs

[Admiral Regulus 2,769]

Oh god, not me again...

 

Anyway the question this time is that a straight path is to be built from B to the nearest point on the road AC. I have to find the length of the straight path. Here's the diagram. I calculated angle ABC using the sine rule in a previous part of the question:

 

IMG_20150309_194138_1425901329681.jpg

 

I don't remember how to do most of this in any sort of timely manner, mainly because I really, really hated this kind of geometry. But... I can do it in Solidworks.

 

Here's what I did. I drew the shape with the given constraints:

 

 

And then I filled in the other dimensions.

 

 

There you go. =D

Edited March 9, 2015 by Admiral Regulus

[IceBeerG 55]

I don't remember how to do most of this in any sort of timely manner, mainly because I really, really hated this kind of geometry. But... I can do it in Solidworks.

 

Here's what I did. I drew the shape with the given constraints:

 

 

And then I filled in the other dimensions.

 

 

There you go. =D

 

That is quite interesting, but may i ask you how you did all these calculations?

 

I can see how it works for an triangle below but not with ABC :/

 

And yeah, with this angle (which seems totally fine with me, even thou i have no idea how admiral regulus found it) h is going to be 79*~0.38/~0,44 =>68.22 (BC) and h will be ~30,01

[Admiral Regulus 2,769]

That is quite interesting, but may i ask you how you did all these calculations?

 

I can see how it works for an triangle below but not with ABC :/

 

And yeah, with this angle (which seems totally fine with me, even thou i have no idea how admiral regulus found it) h is going to be 79*~0.38/~0,44 =>68.22 (BC) and h will be ~30,01

I actually didn't calculate anything. I just drew the figure as a sketch in Solidworks.

 

The software knows the dimensions of the given geometry with the input I gave in the first picture. I modeled the shape, and then asked the software to tell me the unknowns. This is like the digital equivalent of physically drawing the figure and then using a ruler + protractor to measure the unknown angles and lengths.

Edited March 9, 2015 by Admiral Regulus

[IceBeerG 55]

I actually didn't calculate anything. I just drew the figure as a sketch in Solidworks.

 

The software knows the dimensions of the given geometry with the input I gave in the first picture. I modeled the shape, and then asked the software to tell me the unknowns. This is like the digital equivalent of physically drawing the figure and then using a ruler + protractor to measure the unknown angles and lengths.

Oh, but that is too boring =)

 

If someone will actually find these numbers, please share the method

[Admiral Regulus 2,769]

Fine. I'll do it in a little bit. I'm cooking lasagna and typing this from my phone at the moment.

[Taviscratch 453]

Is this thread still a thing?! Well, I'm currently in my last year of studying for an English degree so if you need help with spelling or grammar then I'm your man. I was awesome at maths in college but that was years ago now. Aside from that, I have tomes on philosophy and critics and the like. If you need a quote to prove your point, you come to me.

[Brony Number 42 10,069]

calculated angle ABC using the sine rule in a previous part of the question:

Using the law of sines to find angle CBA

 sin(26) / 79 = sin CBA / 135

CBA = arcsin ( 135*sin(26) / 79) = 48.5138

However, the range of the arcsin function is -pi/2 to pi/2  or -90 to 90 degrees. The angle you need is obtuse, so it sits on the other side of the first maximum of sine. So the real angle is

CBA = 180 - 48.5138 = 131.486

 

angle BAC = 22.514

 

Draw the perpendicular bisector from B to AC (like you did) and label it d. Use the relation sin(22.514) = d / 79

d = 79*sin(22.514) = 30.250

 

The tricky part is remembering the range of arcsin. 

 

 

 

Here's what I did.

Thanks. That's a nice tool to use.

Edited March 9, 2015 by BronyNumber42

[gone~ 614]

Using the law of sines to find angle CBA

 sin(26) / 79 = sin CBA / 135

CBA = arcsin ( 135*sin(26) / 79) = 48.5138

However, the range of the arcsin function is -pi/2 to pi/2  or -90 to 90 degrees. The angle you need is obtuse, so it sits on the other side of the first maximum of sine. So the real angle is

CBA = 180 - 48.5138 = 131.486

 

angle BAC = 22.514

 

Draw the perpendicular bisector from B to AC (like you did) and label it d. Use the relation sin(22.514) = d / 79

d = 79*sin(22.514) = 30.250

 

The tricky part is remembering the range of arcsin. 

 

Rounded to 3 decimal places, the answer isn't 30.250, it's 30.411. ^^

Since AC isn't exactly 135.

 

Although this probably doesn't matter anyways since the question most likely asks for the nearest whole number. ^^;;

[Brony Number 42 10,069]

 

 

Since AC isn't exactly 135.

Yes, you are right. I had 135.263 but I used their rounded number. 

[Starlight Fan 538]

Oh oh, i love mathematic tasks. Back in school i was really good with thous, but now have to look after formula again but... still like to solve them ^^ 

 

First - i doubt that ABC = 54. Can i look at your calculations about this one? (just curious ><)

 

Try this one - S of ABC = 1/2 AC*h. However S of ABC also = 1/2*BC*AC sin BCA. It leads to: 1/2 135*h=1/2*BC*135*~0,44 => h=BC*~0,44, The problem is to fin BC , that can be easily found with angle BAC (in your case is 100) but, again i doubt it is right. If it is, then => BC/sin100=79/sin29=>BC= 79*~0.98/~0,48=>161.3*~0,44 h=>~70,97 (However i find quite strange that there is no absolute number. It's a bit unfamiliar to operate for me. May i ask you what your age, or, better, what is your grade? collage 1st,2nd,3rd? uni? school one of last classes?)

 

I tried to calculate BC somehow negating the fact that ABC is 54, but found this task either to hard to solve for a break or impossible with given information.

 

 

From what I can gather to figure out that question we would need to know either the angles formed when the new line hits the line AC, where it hits the line AC so as to figure out the length of the two sides it makes or the exact split of the angle it makes at B so as to actually know what angles are formed. If we had either of those 3 it could be done easily.

Also angle B is 48.51 I think

 

 

I don't remember how to do most of this in any sort of timely manner, mainly because I really, really hated this kind of geometry. But... I can do it in Solidworks.

 

Here's what I did. I drew the shape with the given constraints:

 

 

And then I filled in the other dimensions.

 

 

There you go. =D

 

 

Using the law of sines to find angle CBA

 sin(26) / 79 = sin CBA / 135

CBA = arcsin ( 135*sin(26) / 79) = 48.5138

However, the range of the arcsin function is -pi/2 to pi/2  or -90 to 90 degrees. The angle you need is obtuse, so it sits on the other side of the first maximum of sine. So the real angle is

CBA = 180 - 48.5138 = 131.486

 

angle BAC = 22.514

 

Draw the perpendicular bisector from B to AC (like you did) and label it d. Use the relation sin(22.514) = d / 79

d = 79*sin(22.514) = 30.250

 

The tricky part is remembering the range of arcsin. 

 

 

 

Thanks. That's a nice tool to use.

 

Thanks for all the answers! I don't have the paper with me at this very moment, but I'll look at all of them closely when I get home. You've all been such a help, I'll thank you all when I graduate... or something like that... or I'll send you e-cards on your birthdays. (Close enough!)

[IceBeerG 55]

Using the law of sines to find angle CBA

 sin(26) / 79 = sin CBA / 135

CBA = arcsin ( 135*sin(26) / 79) = 48.5138

However, the range of the arcsin function is -pi/2 to pi/2  or -90 to 90 degrees. The angle you need is obtuse, so it sits on the other side of the first maximum of sine. So the real angle is

CBA = 180 - 48.5138 = 131.486

 

angle BAC = 22.514

 

Draw the perpendicular bisector from B to AC (like you did) and label it d. Use the relation sin(22.514) = d / 79

d = 79*sin(22.514) = 30.250

 

The tricky part is remembering the range of arcsin. 

 

 

 

Thanks. That's a nice tool to use.

 

Oh right, that rule, yeah, it works =) Thank you =)

[Sommar 1,289]

If anyone needs help on WWI/WWII or Eurpoean history, feel free to post here, and I'll gladly help!(:

[Admiral Regulus 2,769]

Let's use this as a reference. All angles are in green, all lengths are in red.

 

 

All angles in a quadrilateral add up to 360 degrees, so...

 

EQ1:

α + ρ + λ + θ + 26 + β + 77 = 360

α + ρ + λ + θ + β  = 257

 

All angles in a triangle add up to 180 degrees.

 

EQ2:

ρ + λ + 90 = 180

ρ + λ = 90

 

EQ3:

θ + 90 + 26 = 180

θ = 64

 

EQ4:

α + β + 77 = 180

α + β = 103

 

Using the law of cosines:

 

EQ5:

(z + x)² = 120² + 95² - 2*(120)(95)cos(77)

(z + x) = 135.26...

 

Using the law of sines:

 

EQ6:

sin(26) / 79 = sin(λ + θ) / (z + x)

 

Since we know (z + x) and θ, we can solve for λ.

 

sin(26) / 79 = sin(λ + 64) / 135.26...

λ = 67.536...

 

Now, this becomes much easier. There are now four unknowns and four equations, so all of the above angles are solvable.

 

For the actual question, this is just simple trigonometry to find the length w.

 

w = 79*cos(λ)

w = 79*cos(67.36...)

w = 30.41...

 

Finally. That's over with.

[IceBeerG 55]

Let's use this as a reference. All angles are in green, all lengths are in red.

 

 

All angles in a quadrilateral add up to 360 degrees, so...

 

EQ1:

α + ρ + λ + θ + 26 + β + 77 = 360

α + ρ + λ + θ + β  = 257

 

All angles in a triangle add up to 180 degrees.

 

EQ2:

ρ + λ + 90 = 180

ρ + λ = 90

 

EQ3:

θ + 90 + 26 = 180

θ = 64

 

EQ4:

α + β + 77 = 180

α + β = 103

 

Using the law of cosines:

 

EQ5:

(z + x)² = 120² + 95² - 2*(120)(95)cos(77)

(z + x) = 135.26...

 

Using the law of sines:

 

EQ6:

sin(26) / 79 = sin(λ + θ) / (z + x)

 

Since we know (z + x) and θ, we can solve for λ.

 

sin(26) / 79 = sin(λ + 64) / 135.26...

λ = 67.536...

 

Now, this becomes much easier. There are now four unknowns and four equations, so all of the above angles are solvable.

 

For the actual question, this is just simple trigonometry to find the length w.

 

w = 79*cos(λ)

w = 79*cos(67.36...)

w = 30.41...

 

Finally. That's over with.

Finally? That was so much fun to do

[MLPFanatic34 3,234]

If anyone needs help with homework I'd be happy to help, I've been in school and am 24 years old so I know a lot of school subjects.

[Admiral Regulus 2,769]

Finally? That was so much fun to do

 

Not really.

 

I absolutely detest doing tedious things that could otherwise be done by a computer.

[Brony Number 42 10,069]

Here is one from Calculus, Larson and Edwards 10 ed. Consider a polynomial f(x) with real coefficients having the property f(g(x)) = g(f(x)) for every polynomial g(x) with real coefficients. Determine and prove the nature of f(x)

[The Leafeon Pinkeh 3,603]

Not really.

 

I absolutely detest doing tedious things that could otherwise be done by a computer.

But thank you xD I kind of understood that and I have calculus and trig next year. I might do well next year o: even though the teacher is horrible Dx

[Admiral Regulus 2,769]

But thank you xD I kind of understood that and I have calculus and trig next year. I might do well next year o: even though the teacher is horrible Dx

Trig and calculus 1 are easy. For trig, you just have to know the unit circle and trig identities. For calculus 1, the most difficult thing you'll do is integration by substitution.

 

Calculus 2 is hard, though. Calculus 3 is easier, but more tedious. Differential Equations is mind-blowing.

[DATA EXPUNGED 324]

I doubt if anyone here will need help with Vector or Tensor Calculus, but hey, you never know. I guess i could help with Physics, Chemistry, Calculus, Algebra, or any other math related subject. (if you haven't already guessed, I LOVE math)