I thought it was similar triangles

[Nelo]

Homework Statement

http://i54.tinypic.com/2cdib6t.jpg

Since its trig, here is a scan.

Homework Equations

The Attempt at a Solution

I thought it was similar triangles, so i used ratio to solve it as 10.4.

what if it asks for me ar angle instead of a side?

is that correct? The answer key in the back has this page ripped out so I have no idea.

If i am incorrect, may someone guide me in the correct direction?

 

[Mark44]

Homework Statement

http://i54.tinypic.com/2cdib6t.jpg

Since its trig, here is a scan.

Homework Equations

The Attempt at a Solution

I thought it was similar triangles, so i used ratio to solve it as 10.4.

The triangles are similar, but you might not be identifying the corresponding sides or angles correctly. For example, angle BCA = angle DEC.

For side DE, I'm getting 9.7 (rounded to 1 dec. place).

what if it asks for me ar angle instead of a side?

You can use inverse trig functions to find the missing angles.
cos(angle BCA) = 7.5/11.1, so angle BCA = cos^-1(7.5/11.1) \approx 47.5 deg.

is that correct? The answer key in the back has this page ripped out so I have no idea.

If i am incorrect, may someone guide me in the correct direction?

 

[Nelo]

http://i54.tinypic.com/8zk47m.jpg

heres something of the same concept... how do i solve that? there is no info given on the right side of the similar triangle I solved AB to be 21.5, <C is 48 and <A is 42 (degrees)

I don't udnerstand what the acute triangle in the middle has to do with anything, or how to solve theta using ratio comparisons. ANyone?

 

[Nelo]

...?

 

[Mark44]

http://i54.tinypic.com/8zk47m.jpg

heres something of the same concept... how do i solve that? there is no info given on the right side of the similar triangle I solved AB to be 21.5, <C is 48 and <A is 42 (degrees)

I don't udnerstand what the acute triangle in the middle has to do with anything, or how to solve theta using ratio comparisons. ANyone?

The two triangles here (problem #4, I assume) aren't similar, so this is different from the other problem. Here's what you need to do:
1) Find \angle BCA, using exactly the same technique as in the first problem in this thread.
2) When you know \angle BCA, you can find \angle DCE, since \angleBCA + 91.7 deg + \angleDCE = 180 deg.
3) When you have found \angle DCE, \angle \theta = 90 deg - \angle DCE.

 

[Nelo]

I found <C to be 48 degrees
<B is 90 degrees
<A is 42 degrees. all of those added together is not 180, I don't understand what your saying

I got the angles by just doing the inverse and dividing using the proper ratios etc.. idk how to get CDE

 

[Nelo]

Well, i tried the third problem as wel... I am having 0 luck and am completely unable to solve any of these. I don't know how to do 4a, 4b, i got 4c.. can anyone help at all? I just need to know how you get the solution

 

[Mark44]

I found <C to be 48 degrees

Which angle is <C? If you mean <BCA, I get something a bit smaller. The instructions asked for the angles to the nearest tenth of a degree.

<B is 90 degrees

You don't need to find that - it's given.

<A is 42 degrees. all of those added together is not 180, I don't understand what your saying

They're not asking for angle A. What they want is \angle \theta, the angle in the upper right corner.

I got the angles by just doing the inverse and dividing using the proper ratios etc.. idk how to get CDE

See steps 2 and 3 in post #5.

 

[Nelo]

whats the difference between <B and <BCA ? wat , you can't use similar triangles b/c there's no values on the right side

.. to get the angles I am just doing ratios..

ie) <C on the abc side is adjacent/hypotenence which is inverse of cos (19.4/28.8) giving me that value...

Then set theta on a and do opposite over hypoteneuse and do sin inverse 19.4/28.8 etc

 

[Nelo]

derp

 

[Mark44]

The two triangles here (problem #4, I assume) aren't similar, so this is different from the other problem. Here's what you need to do:
1) Find \angle BCA, using exactly the same technique as in the first problem in this thread.
2) When you know \angle BCA, you can find \angle DCE, since \angleBCA + 91.7 deg + \angleDCE = 180 deg.
3) When you have found \angle DCE, \angle \theta = 90 deg - \angle DCE.

whats the difference between <B and <BCA ? wat , you can't use similar triangles b/c there's no values on the right side

\angle B is the right angle at the lower left. \angle BCA is the leftmost of the three angles at the center.

You can't use similar triangles because the two triangles aren't similar.

 

[Nelo]

What does the F pattern and the Z pattern mean on triangles?

 

[Mark44]

ie) <C on the abc side is adjacent/hypotenence which is inverse of cos (19.4/28.8) giving me that value...

I get 47.654°, which rounds to 47.7°.

Then set theta on a and do opposite over hypoteneuse and do sin inverse 19.4/28.8 etc

No, this is wrong. First, you don't set theta - you find it. Second, you can't use inverse sine because the triangles aren't similar.

 

[Mark44]

What does the F pattern and the Z pattern mean on triangles?

I don't know what you mean.

 

[Nelo]

I get 47.654°, which rounds to 47.7°.
No, this is wrong. First, you don't set theta - you find it. Second, you can't use inverse sine because the triangles aren't similar.

?? Its a 90 degree triangle, you can set an angle as the one that you want to find and use trig ratios and inverse(division) to get the angle... Of course you can use inverse

 

[Mark44]

?? Its a 90 degree triangle, you can set an angle as the one that you want to find and use trig ratios and inverse(division) to get the angle... Of course you can use inverse

The only thing you know about the triangle on the right is that it has a 90 degree angle. You DO NOT KNOW the other two angles and you don't know any of the sides.

You seem to be thinking that this problem is similar to the first one in this thread, but it isn't. In the first problem, angle ACE in the middle is given as 90 degrees. That fact, plus the fact that the two triangles are right triangles, makes the triangles similar.

In this problem, angle ACE in the middle is given as 91.7 degrees. That means that the two triangles ARE NOT SIMILAR, so what you propose above WILL NOT WORK.

 

[Nelo]

k... well about this one

http://i54.tinypic.com/232y69.jpg

On the top, 5c) I can't get any of these goddamn questions right and have no script to follow... and I am running out of time

for c) fix i found the middle height by using the sine law which ended up being 61.52. Then I have no idea what to do. I try using everything and nothing seems to work. any help?

 

[Mark44]

k... well about this one

http://i54.tinypic.com/232y69.jpg

On the top, 5c) I can't get any of these goddamn questions right and have no script to follow... and I am running out of time

for c) fix i found the middle height by using the sine law which ended up being 61.52. Then I have no idea what to do. I try using everything and nothing seems to work. any help?

I didn't check, but your number seems about right. Before going any further, let's identify the vertices of the two triangles. Call A the top of the vertical segment, and B its bottom. Call C the left vertex, and D the right vertex.

So you found AB = 61.52 cm.

Your goal is to find \theta, which is \angle BAD.

You know AB and AD, and \angle BDA, so you can use the Law of Sines to find \angle ABD. This is a little tricky, because when you use the inverse sine to find the angle, it will give you an angle that is less than 90 degrees. \angle ABD is clearly larger than 90 degrees, so you'll need to use the idea that sin(90 - x) = sin(90 + x).

For example, sin(60deg) = 1/2, and sin(120deg) = 1/2.
60 = 90 - 30, and 120 = 90 + 30.

 

[Nelo]

nvm i got that one I am stuck on another one ill post it

 

[Nelo]

its not scanned properly but if you can see d) , you can get the correct shape and such . Its split into two and goes down in a diagnol. the end of hte top half is where theta exists. The arrows you can see, and the bottom of the bottom triangle has a angle of 104 degrees and a length of 45m, how do i solve that one? i used cosline law to get the top side which coincides with both triangles, but hteres no angles or anything to solve for theta with

 

[Mark44]

I can't see d well enough to understand what is given. Can you post a clearer scan of it?

 

[Nelo]

http://i52.tinypic.com/653as8.jpg

 

[Mark44]

Let's identify vertices.
Starting at the top left.
Label the left side, from top to bottom, as A, B, and C.
Label the right vertex as D.

Given info:
AB = 10 m.
BC = 20 m.
\angle ACD = 104°

You need to find θ, or \angle BDA.

1. Use the Law of Cosines to find BD.
2. Use the Law of Sines to find \angle CBD.
3. Use the result of step 2 to find \angle ABD - the two angles have to add to 180°.
4. Use the Law of Sines to find AD.
5. Use the Law of Sines or the Law of Cosines to find \angle BDA (θ).

Part of these kinds of problems is figuring out what information you have, and what information you need in order to get to where you want to go. It helps to lay out a plan of how you can get where you need to go.

 

[Nelo]

i get <BD as 53.48, and i solved <CBD , There is still no angle on the top part though, i still can't evaluate theta without having <A / <B in <ABD

 

[Mark44]

It's not <BD; it's just BD, the length of the long side in the smaller triangle. Anyway, I get 53.48 m. for that side, too.

I figured out a slightly simpler way to go.

2. Use the Law of Cosines again to get AD, the long side on the larger triangle. (I get 59.82 m.)
3. Since you now have all three sides of the smaller triangle, use the Law of Sines to get \angle CDB, which we can call \angle \alpha.
4. Since you also have all three sides of the larger triangle, use the Law of Sines again to find \angle CDA, which is the overall angle at the right vertex.
5. Find \angle \theta, which is \angle CDA - \angle \alpha.