I've placed a 2 5 m ladder on a vertical wall such that its height on the wall H = 1 5 m and its distance from the wall D = 2 0 m .
Now, doubling the angle x between the ladder and the ground, I find that the new height H ′ and new distance D ′ are both integer lengths (in meters) as well!
What is H ′ + D ′ ?
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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 , 2 4 , 2 5 ) , so the answer is 7 + 2 4 = 3 1 .
( 7 , 2 4 , 2 5 ) can be generated by noting that 2 5 = 2 7 2 + 1 . Then, the second leg is 2 7 2 − 1 = 2 4 .
tan ( x ) = 2 0 1 5 = 4 3
Using trigonometric formula tan ( 2 β ) = 1 − tan 2 ( β ) 2 tan ( β )
tan ( 2 x ) = 1 − 1 6 9 2 × 4 3 = 7 2 4
Therefore, H ′ = 2 4 k , D ′ = 7 k
Now, ( 2 4 k ) 2 + ( 7 k ) 2 = ( 2 5 ) 2 ⇒ 6 2 5 k 2 = 6 2 5 ⇒ k = 1
Therefore, H ′ + D ′ = 2 4 ( 1 ) + 7 ( 1 ) = 3 1
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?
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For the first angle x , sin x = 2 5 1 5 = 5 3 and cos x = 2 5 2 0 = 5 4 .
For the second angle 2 x , H ′ = 2 5 sin 2 x = 2 5 ⋅ 2 sin x cos x = 2 5 ⋅ 2 ⋅ 5 3 ⋅ 5 4 = 2 4 , and D ′ = 2 5 cos 2 x = 2 5 ( cos 2 x − sin 2 x ) = 2 5 ( ( 5 4 ) 2 − ( 5 3 ) 2 ) = 7 .
Therefore, H ′ + D ′ = 2 4 + 7 = 3 1 .