Height of pole

**aeroflix** · August 24th 2010, 04:15 AM

a pole cast a shadow 15m long when the angle of elevation of the sun is 61 degrees. if the pole is leaned 15 degrees from the vertical directly towards the sun, determine the length of the pole.

**sa-ri-ga-ma** · August 24th 2010, 04:33 AM

Height of the tip of the pole from the ground is Lsinθ. Here θ is ( 90-15) degrees.

Its projection on the ground is x = Lcosθ.

Now tan61 = Lsinθ/(Lcosθ + 15)

Solve the equation and find L.

**Soroban** · August 24th 2010, 07:22 AM

Hello, aeroflix!

The wording of the problem is confusing.
I'll rewrite it to my interpretation.

A pole leans $15^o$ from the vertical directly towards the sun.
The angle of elevation of the sun is $61^o.$
The pole casts a 15-meter shadow on the ground.

Determine the length of the pole.

Code:

      A o
         \  *
          \     *
           \        *
          L \   |       *
             \15|           *
              \ |               *
               \|               61  *
              B o - - - - - - - - - - - o C
                           15

The pole is: $L = AB.$

The shadow is: $BC = 15$ m.

$\angle C = 61^o,\;\;\angle ABC = 105^o$

Hence: . $\angle A \;=\;180^o - 61^o - 105^o \;=\;14^o$

Law of Sines: . $\dfrac{L}{\sin61^o} \:=\:\dfrac{15}{\sin14^o} \quad\Rightarrow\quad L \:=\:\dfrac{15\sin61^o}{\sin14^o}$

Therefore: . $L \;=\;54.22946763 \;\approx\;54.23$ m

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