Trig Story Problems Please Help tonight

[boston1234]

A monument consists of a flagpole 15 m tall standing on a mound in the shape of a cone with verext 140 degrees. How long a shadow does the pole cast on the cone when the angle of elevation of the sun is 62 degrees?
I don't understand how to do this one at all if you could supply a picture that would be great!. Thanks

From the top of an observation post that is 90 m high a ranger sights a campsite at an angle of depression of 10. Turning in a different direction, the ranger sees another campsite at an angle of depression of 13 degrees. The angle between these two lines of sight is 35 degrees . How far apart are the campsites?
This one I attempted and I split them into 2 triangles: the 1st triangle has one side of 90 with angles 17.5, 13, and 149.5, I did the law of sines and did sin 13/90 =sin17.5/x and got 120 m and the second triangle I got one side of 90 with angles 17.5, 10, and 152.5 and used law of sines and did sin 10/90 = sin 17.5/x and got 156m so my final answer was 276 m is that correct?

 

[soroban]

Hello, boston1234!

Here's the first one . . .

A monument consists of a flagpole 15 m tall standing on a mound
in the shape of a cone with vertex angle 140 degrees.
How long a shadow does the pole cast on the cone
when the angle of elevation of the sun is 62 degrees?

                          o A
                        * |
                      * 28| 15
                    *     |
                  * 62    |
              D o - - - - o B
              *       * 70:   *
            *     *       :       *
          * 42*           :           *
        * *               :               *
    C o                   :                   *
  * . . - - - - - - - - - + - - - - - - - - - - - *
                          E

\(\displaystyle \text{The flagpole is: }\:AB = 15\)

\(\displaystyle \text{The shadow is }BC.\)

\(\displaystyle BD\text{ is horizontal.}\)

\(\displaystyle \text{At the vertex of the cone: }\:\angle CBE = 70^o \quad\Rightarrow\quad \angle DBC = 20^o\)
. . \(\displaystyle \text{Then: }\:\angle ABC = 110^o,\)

\(\displaystyle \text{The angle of elevation of the sun is: }\:\angle ADB = 62^o.\)
. . \(\displaystyle \text{Then: }\:\angle A = 28^o.\)

\(\displaystyle \text{In }\Delta ABC\!:\;\angle C \;=\;180^o - 110^o - 28^o \:=\:42^o\)


\(\displaystyle \text{Law of Sines: }\;\frac{BC}{\sin28^o} \:=\:\frac{15}{\sin42^o}\)

Got it?

 

[soroban]

Hello again, boston1234!

I don't know how you found those strange angles . . .

From the top of an observation post that is 90 m high a ranger sights a campsite at an angle of depression of 10.
Turning in a different direction, the ranger sees another campsite at an angle of depression of 13 degrees.
The angle between these two lines of sight is 35 degrees.
How far apart are the campsites?

\(\displaystyle \text{The tower is: }\:AB = 90\)

\(\displaystyle \text{The ranger looks in the first direction and sights campsite }C.\)

      - - - - - - * A
            10  * |
              * 80|
            *     | 90
          *       |
        *         |
    C * - - - - - * B

\(\displaystyle \text{The angle of depression is }10^o \quad\Rightarrow\quad \angle CAB = 80^o\)

\(\displaystyle \text{We have: }\:\tan80^o \:=\:\frac{BC}{90} \quad\Rightarrow\quad BC \:=\:90\tan80^o \;\approx\;510.4\text{ m.}\)



\(\displaystyle \text{He looks in a second direction and sights campsite }D.\)

    A * - - - - -
      | * 13
      |77 *
   90 |     *
      |       *
      |         *
    B * - - - - - * D

\(\displaystyle \text{The angle of depression is }13^o \quad\Rightarrow\quad \angle DAB = 77^o\)

\(\displaystyle \text{We have: }\:\tan77^o \:=\:\frac{BD}{90} \quad\Rightarrow\quad BD \:=\:90\tan77^o \;\approx\;389.8\text{ m.}\)


\(\displaystyle \text{Looking down at the ground, we have this diagram:}\)

            B
            o
           *35* 
          *     *  389.8
         *        *
 510.4  *           o D
       *        *
      *     *
     *  *
  C *

\(\displaystyle \text{Use the Law of Cosines: }\;CD^2 \;=\;510.4^2 + 389.8^2 - 2(510.4)(389.8)\cos35^o\)

Got it?