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Hello, this is the answer of the first question of the Eiffel tower's problem.
We can't do the others questions because the picture is too little. Can you send
a bigger please ? Thank you.

ABC is isoscele in C.
We call [AH] the leaving height of A.
[AH] divide BA in two equal segments : CH= c×1/2= 97,5×1/2= 48,75 m
[AH] is perpendicular to (AB) so AHC is a right-angle triangle.
We can use the propriety of Pythagore :
HC² + AH² = CA²
48,5² + 800² = CA²
2376,2 + 640000 = CA²
642376,2 = CA²
The 642376,2's square root is equal to 801,5
CA = b = a = 801,5m

Now, we can use trigonometry:
cos(Ĉ) = 800/801,5
Ĉ = cos-1×800/801,5  3,5°
Cos(Â) = 48,5/801,5
 = cos-1×48,5/801,5  86,5°

Posted in Group3.

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