Could someone help me on these problems?

Hello, hamster!

There must be more information . . .
. . as stated, the problems do not have unique answers.

Are any of the lines perpendicular?
Are any triangles isosceles?
 
It might help if you (1) trimmed the image to get rid of the empty half and (2) labelled vertices and specified what numbers went with which bits and whether the numbers are segment lengths or angle measures. For instance:

. . . . .
precxc1.gif


Then clarify what the numbers stand for, using the letters as reference points.

Thank you.

Eliz.
 
Im sorry that is all the information i was given in my worksheet. all the numbers are degrees...no right triangles. its supposed to be figuring out the height of the mock "sears tower" (figure 1) and "empire state" building (figure 2)
 
Hello, hamster!

Im sorry that is all the information i was given in my worksheet.
All the numbers are degrees...no right triangles.
It's supposed to be figuring out the height of the mock "sears tower" (figure 1)
and "empire state" building (figure 2)
Click to expand...
You're wrong! . . . There are right triangles!
    \displaystyle \;\;Why is one angle marked with a black arc?
    \displaystyle \;\;I bet it's a right angle.

And you didn't make it clear (until now) that it's a three-dimensional diagram.
Code: 
   D
    *
    |\
    | \
   h|  \               *C
    |   \       *     *
    |    \*          *
   B*     \         *
      *    \       *
        *   \     *
          *  \   *
            * \ *
               *
                A
See if this makes sense . . .

ΔABC\displaystyle \Delta ABC is flat on the ground.
h = BD\displaystyle h\,=\,BD is the Sears Tower, perpendicular to the ground.
      \displaystyle \;\;\;Hence: ∠DBA = 90o\displaystyle \angle DBA\,=\,90^o

Distance AC = 673\displaystyle AC\,=\,673 feet.  ∠BCA = 49.3o, ∠DAB = 61.6o\displaystyle \:\angle BCA\,=\,49.3^o,\:\angle DAB\,=\,61.6^o

But even that is <u>not</u> enough information for a unique answer.

So I must assume that ∠BAC = 90o\displaystyle \angle BAC\,=\,90^o
    \displaystyle \;\;(Why else would it be marked so clearly?)


Game plan:
In right triangle BAC\displaystyle BAC, find the length of AB\displaystyle AB.
Then in right triangle DBA\displaystyle DBA, find the length of h = BD\displaystyle h\,=\,BD.

I'll let someone else do the work . . . I need a nap.
 
haha could you please continue? i am so lost...thank you for all your help. i apperciate it!
 
hamster said:
haha could you please continue? i am so lost...thank you for all your help. i apperciate it!
Click to expand...

Where are you "so lost?" Soroban has given you the plan to solve this problem. I'll go just a bit further into the plan:

In right triangle BAC, with right angle at A, the two legs are BA and CA, and
tan C = AB/AC

You know that angle C has a measure of 49.3 degrees, and that AC = 673:

tan 49.3 = AB/673

Multiply both sides by 673:

673* tan 49.3 = AB

That will give you the length of leg AB, which is also a leg of right triangle ABD. The length of the other leg DB of this triangle is the height h you are looking for. In this triangle,
tan (angle DAB) = DB/AB

And you are given that the measure of angle DAB is 61.6 degrees, so

tan 61.6 = h/AB

But you will have found AB already, right? So, substitute the length you got for AB, and solve for h......
 
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