I have this problem...i know the answer (from the back of the book) but I have no idea how to solve this....could someone please help me?
Okay, draw a horizontal line from the tip of the shadow, straight left. It should intersect the flagpole. Now you have two 90 degree triangles to work with. One with an angle of 70, and the other with an angle of 76.
What you need to do is find the adjacent side of the angle of 76, and add it to the adjacent side of the angle of 70. You can see that together, they make up the height of the flagpole.
One of our triangles has an angle of 76, and a hypotenuse of 16. You can find the opposite angle because $\displaystyle sin(76)=\frac{opp}{16} \to 16sin(76)=opp$ And you can find the adjacent side by the same method, with cosine, so the adjacent side is 16cos(76).
Now you know that your opposite side of the angle of 76 is shared as the opposite side of the angle of 70, so now you have an angle and a side. So we want to find the adjacent side, we know our angle and our opposite side, so tangent will give us what we need. $\displaystyle tan(70) = \frac{16sin(76)}{adj} \to adj=\frac{16sin(76)}{tan(70)}$
Now you have equations which give you your adjacent sides for both triangles, just add them together
$\displaystyle 16cos(76) + \frac{16sin(76)}{tan(70)}$
And you can see by plugging it into a calculator that the answer is 9.521290534 meters, so we know by verifying against the back of the book, that we did it correctly.
Hello, aikenfan!
A flagpole is mounted on the front of a library's roof. .From a point 100 feet
in front of the libary, the angles of elevation to the base of the flagpole and
the top of the flagpole are 28° and 30°45', respectively.
. . Find the height of the flagpole.
There are two diagrams to consider:The height of the library is $\displaystyle y.$Code:* * | * |y * 28° | * - - - - - - - - - - - * 100
We have: .$\displaystyle \tan(28^o) \:=\:\frac{y}{100}\quad\Rightarrow\quad y \:=\:100\!\cdot\!\tan(28^o)$ .[1]
The height of the flagpole is $\displaystyle h.$Code:* * | * |h * | * * * | * |y * 30¾° | * - - - - - - - - - - - * 100
We have: .$\displaystyle \tan(30.75^o) \:=\:\frac{h+y}{100}\quad\Rightarrow\quad h \:=\:100\!\cdot\!\tan(30.75^o) - y$ .[2]
Substitute [1] into [2]: .$\displaystyle h \;=\;100\!\cdot\!\tan(30.75^o) - 100\!\cdot\!\tan(28^o)$ . . . There!