triangle

unique · 2006-10-08 08:25:58

solve by drawing the necessary reference triangle and evaluating:

sin 60 degrees + sqrt3 / 2 times sin 30 degrees

i did

sin 60 equals 0.5 > 1/2
sin 30 equals 0.5 > 1/2

1/2 + sqrt3 / 2 * 1/2
sqrt3 /2 times 1/2 equals sqrt3 / 4

1/2 + sqrt3 / 4
add them
LCD is 8
so:

4 + 2 sqrt3
---------------
8

cross out.
4 and 8 goes into 2

2 + 2 sqrt3
--------------
4

cross out
2 + 2sqrt3
--------------
2

cross out 2's
leaving

2sqrt3

now how do you make that into a triangle.
and please do tell me if i did this right. thanks

fgarb · 2006-10-08 18:13:47

First, I should warn you that you made a mistake in your math near the beginning. Make sure that you understand what the proper values are for sin(30), sin(60), cos(30), and cos(60) are. There's a pretty handy way of thinking about this that makes it so you don't have to just memorize the numbers. I'm sure someone on this forum can explain the trick to you if you don't know it, that way you'll be less likely to make such mistakes in the future.

As for the reference triangle, I'm not sure what you're teacher is looking for here (maybe someone else knows). I have a guess, but it's going to be very difficult for me to communicate my idea without being able to draw a picture, so I'm going to try to make a text picture, but it will be difficult to read.

*****Warning: the explanation below is kind of strange. Maybe someone else can do a better job of communicating the idea of this problem. *******

My guess is that your teacher wants you to express everything in terms of one trig function instead of two. Each trig function can be graphically represented as a triangle, so from the original expression, you could graphically represent the answer like this:

/|
sqrt(3)/2 / | (sqrt(3)/2) * sin(30)
/30__|
/| |
/ | | A
1 / |sin(60)
/ | |
/_60____ |

In this picture there are two right triangles with touching corners. The first has angle 60 degrees and hypoteneuse 1, and the second has angle 30 and a hypoteneuse of length sqrt(3)/2 (not drawn to scale, sorry). Then, your answer is the total length of side A, the sum of the lengths of the vertical sides of the two triangles:

A = 1*sin(60) + sqrt(3)/2*sin(30)

But if the angles of the two triangles were the same then you could write this all as one large triangle because there wouldn't be a bend between the two smaller triangles. The trouble is that you have angles 60 and 30. Maybe your teacher wants you to make them both 60. You can use:

sqrt(3)/2 * sin(30) = sin(60)*(1/2) by substituting sin(30) = 1/2 and sqrt(3)/2 = sin(60)

Then you can write

A = 1*sin(60) + 1/2*sin(60) = 3/2 * sin(60)
(you should check to make sure you get the same answer using both approaches to make sure you didn't make any mistakes)

This is equivalent to a triangle:

/|
/ |
3/2/ |
/ | A
/ |
/_60___

This *might* be what your teacher wants you to draw. The question seems a bit vague to me. Maybe someone else has a different idea.

unique · 2006-10-10 01:53:09

are you sure? because i had another problem such as this one
draw a triangle and evaluate 2sqrt(3) * cos60
so
2sqrt(3)cos(60)
2sqrt(3) * 1/2
2sqrt(3) / 2
cross out 2
leaving
sqrt(3) as teh answer
for teh triangle
on the hypot side we put 2
on the opposite side we put sqrt(3)
and adj side we put 1
and for teh angles we had 30dgerees and 60 degrees

so i thought we were supposed to do that here..........

Math Is Fun Forum

#1 2006-10-08 08:25:58

triangle

#2 2006-10-08 18:13:47

Re: triangle

#3 2006-10-10 01:53:09

Re: triangle

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