Trig- triangles

[kel]

A flagpole 4m tall stands on a sloping roof. A guy wire 5m long joins the top of the pole to a point on the roof 6m up from the bottom of the pole. At what angle is the roof inclined to the horizontal?


I need help setting up the problem. Thanks.

 

[tkhunny]

A flagpole 4m tall stands on a sloping roof. A guy wire 5m long joins the top of the pole to a point on the roof 6m up from the bottom of the pole. At what angle is the roof inclined to the horizontal?

Ill-defined question.

1) Is the flagpole vertical? This MAY be a safe assumption.
2) What does "up from the bottom of the pole" mean? Along the roof? Vertical distance?

 

[soroban]

Hello, kel!

This is not an easy set-up . . . it took me three tries to get a "nice" one.

A flagpole 4m tall stands on a sloping roof.
A guy wire 5m long joins the top of the pole to a point on the roof 6m up from the bottom of the pole.
At what angle is the roof inclined to the horizontal?

            A
            *
            | \
           4|   \5                  *F
            |     \         *
            |   6   *C
           B*
    * - - - - - - - - - - - - - - - -
    D                               E

\(\displaystyle DE\) is horizontal, \(\displaystyle DF\) is the slanted roof, \(\displaystyle AB\) is the vertical flagpole (4m).
\(\displaystyle AC\) is the guy wire (5m) , \(\displaystyle BC = 6\)m.

If the roof were horizontal, \(\displaystyle \angle ABC\) would be \(\displaystyle 90^o\).
. . Since the roof is not horizontal, \(\displaystyle \angle ABC\) is somewhat less than \(\displaystyle 90^o\).
. . Very well, how much less?

Using the Law of Cosines: \(\displaystyle \;\cos(\angle ABC) \:=\:\frac{4^2\,+\,6^2\,-\,5^2}{2\cdot4\cdot6}\:=\:\frac{9}{16}\;\;\Rightarrow\;\;\angle ABC\:\approx\:55.8^o\)

Therefore, the roof is slanted at: \(\displaystyle \:90^o\,-\,55.8^o\:=\:34.3^o\)

 

[galactus]

\(\displaystyle sqrt{25-x^{2}}=sqrt{36-(4-x)^{2}}\)

square both sides:

\(\displaystyle 25-x^{2}=36-(4-x)^{2}\)

Simplify:

\(\displaystyle 25-x^{2}-(-x^{2}+8x+20)=5-8x\)

\(\displaystyle x=\frac{5}{8}\)

\(\displaystyle 4-x=\frac{27}{8}=3.375\)

\(\displaystyle A=sin^{-1}(\frac{3.375}{6})=34.23\)degrees

 

[kel]

thanks for your help.