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.