Doubling the angle from the ladder

Geometry Level 2

I've placed a 25 m 25\text{ m} ladder on a vertical wall such that its height on the wall H = 15 m H=15\text{ m} and its distance from the wall D = 20 m . D=20\text{ m}.

Now, doubling the angle x x between the ladder and the ground, I find that the new height H H' and new distance D D' are both integer lengths (in meters) as well!

What is H + D ? H' + D'?


The answer is 31.

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4 solutions

David Vreken
Nov 26, 2018

For the first angle x x , sin x = 15 25 = 3 5 \sin x = \frac{15}{25} = \frac{3}{5} and cos x = 20 25 = 4 5 \cos x = \frac{20}{25} = \frac{4}{5} .

For the second angle 2 x 2x , H = 25 sin 2 x = 25 2 sin x cos x = 25 2 3 5 4 5 = 24 H' = 25 \sin 2x = 25 \cdot 2 \sin x \cos x = 25 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = 24 , and D = 25 cos 2 x = 25 ( cos 2 x sin 2 x ) = 25 ( ( 4 5 ) 2 ( 3 5 ) 2 ) = 7 D' = 25 \cos 2x = 25 (\cos^2 x - \sin^2 x) = 25 ((\frac{4}{5})^2 - (\frac{3}{5})^2) = 7 .

Therefore, H + D = 24 + 7 = 31 H' + D' = 24 + 7 = \boxed{31} .

Henry U
Nov 27, 2018

Since the information is given that all lengths are integers again, we just have to find another pythagorean triple with a hypothenuse of 25. The only possibility is ( 7 , 24 , 25 ) (7, 24, 25) , so the answer is 7 + 24 = 31 7+24 = \boxed{31} .


( 7 , 24 , 25 ) ({\color{#D61F06}7}, 24, 25) can be generated by noting that 25 = 7 2 + 1 2 25 = \frac {{\color{#D61F06}7}^2+1}{2} . Then, the second leg is 7 2 1 2 = 24 \frac {{\color{#D61F06}7}^2-1}{2} = 24 .

tan ( x ) = 15 20 = 3 4 \tan (x) = \dfrac {15}{20} = \dfrac 34

Using trigonometric formula tan ( 2 β ) = 2 tan ( β ) 1 tan 2 ( β ) \tan(2β) = \dfrac {2\tan (β)}{1 - \tan^2(β)}

tan ( 2 x ) = 2 × 3 4 1 9 16 = 24 7 \tan (2x) = \dfrac {2\times \frac 34}{1 - \frac {9}{16}} = \dfrac {24}{7}

Therefore, H = 24 k , D = 7 k H'= 24k , D'= 7k

Now, ( 24 k ) 2 + ( 7 k ) 2 = ( 25 ) 2 625 k 2 = 625 k = 1 (24k)^2 + (7k)^2 = (25)^2 \Rightarrow 625 k^2 = 625 \Rightarrow k = 1

Therefore, H + D = 24 ( 1 ) + 7 ( 1 ) = 31 H'+ D'= 24(1) + 7(1) =\boxed{31}

Roger Erisman
Dec 21, 2018

25^2 = 625 so need to find 2 integers whose squares sum to 625.

Largest must be less than 25.

24^2 = 576.

625 - 576 = 49.

Sqrt. of 49 is 7.

24 + 7 = 31.

While you have certainly found the correct values, is this the only solution? Why or why not?

Pi Han Goh - 2 years, 5 months ago

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