1. ## find missing angles...?

hi, i need to find the missing angles from this shape:

i know it has something to do with sin, cos and tan.
im not really sure what to do, i think you would use cos because the shape has adjacent and hype? apart from that i can't really remember what to do! so could someone please help me?

Thanks

2. It is easy to find the missing angle since a triangle has 180 degrees and you know two of the angles. The missing angle is 90 - 57 = 33 degrees.

3. i need to find what the lin on the left is with the question mark

4. Now you know the missing angle is 33 degrees you can use the sine rule:

$\displaystyle \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ where side a is opposite angle A and the same for B and C

In other words $\displaystyle \dfrac{a}{\sin 33^{\circ}} = \dfrac{11}{\sin 90^{\circ}}$ and since $\displaystyle \sin 90 = 1$ then you can solve for the missing side a:

edit: the title is misleading if you want to find out the side...

5. Originally Posted by e^(i*pi)
Now you know the missing angle is 33 degrees you can use the sine rule:

$\displaystyle \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ where side a is opposite angle A and the same for B and C

In other words $\displaystyle \dfrac{a}{\sin 33^{\circ}} = \dfrac{11}{\sin 90^{\circ}}$ and since $\displaystyle \sin 90 = 1$ then you can solve for the missing side a:

edit: the title is misleading if you want to find out the side...
hi, sorry for the late response, had to pop out to get some christmas presents!

so, carrying it on from what you posted i would do, 11xsin33/sin90= 6?

i also relised my mistake while walking back home! i had wrote down the question wrong in my book!

6. You can also just use the fact that the cosine of an angle is adjacent/hypotenuse:

cos 57 = x/11. So x = 11cos 57

7. Originally Posted by DrSteve
You can also just use the fact that the cosine of an angle is adjacent/hypotenuse:

cos 57 = x/11. So x = 11cos 57
Hi, thanks for posting! But when I did it the way e^(i*pi) showed me I got a different answer to what I got when I did it your way?!? When I did it e^(i*pi)s way I got the answer of 6 but for your way I got 57.98. Which way is right?

8. Originally Posted by andyboy179
Hi, thanks for posting! But when I did it the way e^(i*pi) showed me I got a different answer to what I got when I did it your way?!? When I did it e^(i*pi)s way I got the answer of 6 but for your way I got 57.98. Which way is right?
Both ways are correct, I checked Dr Steve's method and got 5.99 [only put 6 if you were instructed to round to the nearest whole number].

Is your calculator in degrees mode?

Edit: if you think about it $\displaystyle |\cos(x)| \leq 1$ so the answer won't be more than 11 for sure. I'm confused as to how you get 57.98...

9. So what I done on my earlier response to your post was correct?
To get 57.98 I did 11cos + 57.

10. Post 5 is fine.

What Dr Steve meant is that you should take the cosine of 57. No plus sign involved (and you can't take the cos of nothing, I can't believe your calculator let you either)

$\displaystyle x = 11\cos(57^{\circ}) \approx 5.99$

11. Oh ok, thanks guys!

12. Originally Posted by andyboy179
So what I done on my earlier response to your post was correct?
To get 57.98 I did 11cos + 57.
If you enter 11 on your calculator and then press the cos button, you'll get $\displaystyle \cos(11^\circ)$, which is not what you wanted. You need to get $\displaystyle \cos(57^\circ)$ first, then multiply it by 11.

13. Math has many difficult concepts that are not easy to solve...i have to solve many problems while solving geometry equations like finding angles ...i am looking for the well qualified math tutor that is able to solve all concepts.

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