Question #06927

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Question #06927

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Answer 1 · 2017-04-06T19:44:06

I found $195.7 m$ $195.7m$ but I am not really sure of my method. Anyway check it and also check my maths!!!

Explanation:

Ok it is in Geometry but I had to use trignometry as well.
The object is a kind of wedge and I used this picture as well:
enter image source here
I first consider the two triangles ( I assume them right triangles in C) CPA and CPB and using trigonometry I say that:
$H = Q sin (57^{\circ})$ $H=Q sin(57^@)$
$H = S sin (31^{\circ})$ $H=S sin(31^@)$
$C A = S cos (31^{\circ})$ $CA= S cos(31^@)$
$C B = Q cos (57^{\circ})$ $CB= Q cos(57^@)$

Where $Q and S$ $Q and S$ are the hipotenuses.
Then I consider triangle ABC (right at B) and apply and Pythagoras Theorem to get:
$C A^{2} = A B^{2} + B C^{2}$ $CA^2=AB^2+BC^2$
or:
$S^{2} {cos}^{2} (31^{\circ}) = 300^{2} + Q^{2} {cos}^{2} (57^{\circ})$ $S^2 cos^2(31^@)=300^2+Q^2 cos^2(57^@)$ (1)
BUT:
$H = Q sin (57^{\circ})$ $H=Q sin(57^@)$
$H = S sin (31^{\circ})$ $H=S sin(31^@)$
so that:
$Q = \frac{H}{sin (57^{\circ})}$ $Q=H/ sin(57^@)$
$S = \frac{H}{sin (31^{\circ})}$ $S=H/ sin(31^@)$
that substituted into (1) give:
$\frac{H^{2}}{{sin}^{2} (31^{\circ})} {cos}^{2} (31^{\circ}) = 300^{2} + \frac{H^{2}}{{sin}^{2} (57^{\circ})} {cos}^{2} (57^{\circ})$ $H^2/ sin^2(31^@)cos^2(31^@)=300^2+H^2/ sin^2(57^@) cos^2(57^@)$
rearranging:
$H^{2} [\frac{{cos}^{2} (31^{\circ})}{{sin}^{2} (31^{\circ})} - \frac{{cos}^{2} (57^{\circ})}{{sin}^{2} (57^{\circ})}] = 300^{2}$ $H^2[cos^2(31^@)/sin^2(31^@)- cos^2(57^@)/sin^2(57^@)]=300^2$
doing some calculations I got:
$H^{2} = \frac{300^{2}}{2.348}$ $H^2=300^2/2.348$
$H = 195.7$ $H=195.7$

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Question #06927

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