Question 1025276
your solution is shown below in the following 4 worksheets.


the first worksheet is the diagram i drew of the problem.


the overall diagram is on top.


the separate triangles formed are in the middle and the bottom of the diagram.


the second worksheet shows the derivation of the solutions to your first two questions.


the bottom of the second worksheet and the third worksheet show the derivation of the solution to your third question.


the fourth worksheet confirms that the value of HT is correct whether you use triangle AHT or BHT to confirm.


you should be able to follow the progression.


if you have any questions, send me an emails and i'll explain further.


the solutions take advantage of the fact that tan(45) and tan(30) = opposite side divided by adjacent side.


the third solution takes advantage of the fact that the hypotenuse of a right triangle is equal to the square root of the first leg squared plus the second leg squared.


you first derive HT = w * tan(45).


you next derive BT = HT / tan(30).


because HT is equal to w * tan(45), you substitute for HT to get the final derivation of BT = w * tan(45) / tan(30).


your solution to question number 3 BT squared with its equivalent of (w * tan(45) / tan(30))^2 to arrive at the final solution.


the worksheets are shown below:


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<img src = "http://theo.x10hosting.com/2016/031602.jpg" alt="$$$" </>


<img src = "http://theo.x10hosting.com/2016/031603.jpg" alt="$$$" </>


<img src = "http://theo.x10hosting.com/2016/031604.jpg" alt="$$$" </>