Question 1076426
let T equal the top of the tree and B equal the bottom of the tree.


you have 2 triangles formed with a common side being side TB


the first triangle formed is PTB with angle P equal to 30 degrees.


the second triangle formed is QTB with angle Q eaual to 45 degrees.


both triangle PTB and QTB are part of the overall triangle formed of PTQ.


TB is assumed to be perpendicular to PQ.


therefore triangles PTB and QTB are assumed to be right triangles.


you are given that the length of PQ is equal to 100 meters.


tan(30) = TB / PB


tan(45) = TB / BQ


solve for TB in both these formulas to get:


TB = PB * tan(30)


TB = BQ * tan(45)


since they are both equal to TB, then you get:


PB * tan(30) = BQ * tan(45)


let x equal the length of PB.


therefore 100 - x equals the length of BQ.


the formula becomes:


x * tan(30) = (100 - x) * tan(45)


simplify to get:


x * tan(30) = 100 * tan(45) - tan(45) * x


add tan(45) * x to both sides of this equation to get:


x * tan(30) + x * tan(45) = 100 * tan(45)


factor out the x to get:


x * (tan(30) + tan(45)) = 100 * tan(45)


divide both sides of the equation by (tan(30) + tan(45)) to get:


x = (100 * tan(45) / (tan(30) + tan(45))


solve for x to get:


x = 63.39745962


that's the length of PB.


100 - x = 36.60254038


that's the length of BQ


you were asked to find the distance of p from the tree.


that would be the length of PB.


TB is the height of the tree and that turns out to be 36.60254038 meters.


your diagram would look like this:


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