# Thread: A proof my friend got me to do

1. ## A proof my friend got me to do

My friend and I both like simple calculus, he showed me this proof which looks straightforward however i could not do it. I drew a diagram on paint and it is an attachment. The aim is to show that the height of P above the floor after being tilted is h(cosb+2sinb).

2. from the first picture ...

diagonal length = $\sqrt{5} \cdot h$

$\sin(a) = \dfrac{1}{\sqrt{5}}$

$\cos(a) = \dfrac{2}{\sqrt{5}}$

after tilting at an angle $b$ ...

height = $\sqrt{5} \cdot h \cdot \sin(a+b)
$

height = $\sqrt{5} \cdot h [\sin(a)\cos(b) + \cos(a)\sin(b)]
$

height = $\sqrt{5} \cdot h \left[\dfrac{1}{\sqrt{5}} \cdot \cos(b) + \dfrac{2}{\sqrt{5}} \cdot \sin(b) \right]$

height = $h \left[\cos(b) + 2\sin(b) \right]$

3. ## Can someone please check my proof

I completed this proof in the attachment but can someone please check my working

The aim was to show that the height of P above floor after the crate is tilted is h(cosb+2sinb)

From the rectangles you can work out
Length of OP=h√5
sin(a)=1/√5
cos(a)=2/√5

therefore
height=h√5.sin(a+b)
=h√5.(sinacosb+cosasinb)
=h√5.(1/√5cosb+2/√5sinb)
=h(cosb+2sinb)