Trigonometry World Problem - Angle of Elevation

**magnaye** · July 2nd 2010, 05:26 AM

6. From the top of a building 60 ft. high the angle of elevation of the top of a communications tower is 13.8 degrees . From the bottom of the building the angle of elevation of the top of the tower is 35.2 degrees. Find the Height of the tower. Also, determine the distance from the building to the tower

Ok about the sketch part, I need help on it especially on the part: From the top of a building 60 ft. high the angle of elevation of the top of a communications tower is 13.8 degrees argh its hard to be a math newbie

**Soroban** · July 2nd 2010, 09:12 AM

Hello, magnaye!

6. From the top of a building 60 ft. high the angle of elevation
of the top of a communications tower is $13.8^o.$
From the bottom of the building the angle of elevation
of the top of the tower is $35.2^o.$

Find the Height of the tower.
Also, determine the distance from the building to the tower

Code:

                              * C
                          * * |
                      *   *   |
                  *     *     |
              *       *       |
          * 13.8    *         |
    A * - - - - - * - - - - - | y
      |         *             |
      |       *               |
   60 |     *                 |
      |   *                   |
      | * 35.2                |
    B * - - - - - - - - - - - * D
                  x

Let $x \,=\,BD,\;y \,=\,CD$

Since $\angle CBD = 35.2^o$ , then: $\angle ABC = 54.8^o$

$\angle CAB \:=\:13.8^o + 90^o \:=\:103.8^o$

Hence: . $\angle ACB \:=\:180^o - 54.8^o - 103.8^o \:=\:21.4^o$

In $\Delta ABC$ , Law of Sines: . $\dfrac{BC}{\sin103.8^o} \:=\:\dfrac{60}{\sin21.4^o}$

. . $BC \:=\:\dfrac{60\sin103.8^o}{\sin21.4^o} \:=\: 159.6924203 \;\approx\;159.69$

In right triangle $CDB\!:\;\;\sin35.2^o \:=\:\dfrac{y}{BC}<br />$

. . $y \:=\:159.69\sin35.2^o \:=\:92.05047657$

Therefore: . $\boxed{y \;\approx\;92.05\text{ ft}}$

In right triangle $CDB\!:\;\;\cos38.2^o \:=\:\dfrac{x}{BC}$

. . $x \:=\:159.69\cos35.2^o \:=\:130.4898688$

Therefore: . $\boxed{x \;\approx\;130.49\text{ ft}}$

**magnaye** · July 3rd 2010, 04:12 AM

by many thanks Soroban, I stand correct with my sketch, now I realized that the building is separate with the communications tower

I thought that the communications tower is above the building which is wrong.... Thanks for the enlightenment. and more blessings Soroban cheers

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