Poles in the Field

Geometry Level 3

ABC is a field in the form of an equilateral triangle. Two vertical poles of heights 45m and 20m are erected at A and B respectively. The angles of elevation of the tops of the two poles from C are complementary to each other. There is a point D on AB such that from it, the angles of elevation of the tops of the two poles are equal.

Then the length of AD is (in meters) is [the answer is in mixed fraction]

20\frac { 5 }{ 13 }

20\frac { 10 }{ 13 }

17\frac { 5 }{ 12 }

17\frac { 10 }{ 12 }

1 solution

Roger Erisman
May 23, 2017

Call the side of the equilateral triangle, X and the elevation angle to top of poles angle C.

Then 45 / X = tan C and 20 / X = tan(90 - C)

Tan (90 - C) = Cot C = 1/Tan C

45 / tan C = 20 / (1/tan C) Solving gives C = 56.3 degrees.

X = 45 / tan 56.3 yielding X = 30

Let DA = Y and angle of elevation = D

tan D = 45 / Y and tan D = 20 / (30 - Y)

Cross multiply and solve Y = 20.8

To best visualize the fact that $\angle BCB’$ and $\angle ACA’$ are complementary, I have put them into the same plane, with $BB’$ pointing downward. We can see that $\triangle BCB’$ is similar to $\triangle AA’C$ , so that

$\dfrac {x}{20}=\dfrac{45}{x}$

From which $x=30$

$\triangle DAA’$ is similar to $\triangle DBB’$ with a ratio of sizes being

$\dfrac{20}{45}=\dfrac94$

If we divide $x=30$ in the ratio $\dfrac94$ , we get $AD=\dfrac{270}{13}$

Writing it as a mixed fraction it will become $\boxed{20\dfrac{10}{13}}$

Marta Reece - 4 years ago

Poles in the Field

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