Question 1

Miki is standing some distance away from a building. She measures the angle of elevation to the top of the building to be 20 degrees. She then moves 10m closer to the building and finds the new angle of elevation to the top of the building to be 35 degrees. (all answers should be in 3 significant figures).a. Determine the two different equations necessary to find x and h.b. Determine the height, h , of the building.

Miki is standing some distance away from a building. She measures the angle of elevation-example-1

Question 2

$\begin{gathered} \text{From the traingle, B} \\ U\text{sing pythgoras theorem} \\ \text{Opp = h} \\ \text{Adj = x} \\ \theta\text{ = 35} \\ \tan \theta\text{ = }(h)/(x) \\ \tan \text{ 35 = }(h)/(x) \\ 0.7\text{ = }(h)/(x) \\ By\text{ cross multiplying} \\ 0.7x\text{ = h}-----------------------(1) \\ x\text{ = }(h)/(0.7) \\ \\ To\text{ solve for h} \\ \text{opp = h} \\ \text{Adj = (10 + x)} \\ \theta\text{ = 20} \\ \tan \theta\text{ = }\frac{\text{Opp}}{\text{Adj}} \\ \tan \text{ 20 = }(h)/((10+x)) \\ 0.36\text{ = }(h)/((10+x)) \\ 0.36\text{ = }(h)/((10+x)) \\ h\text{ = 0.36(10+x)}--------------------(1) \\ But\text{ equation 1 and equation 2 are the same since they are both h} \\ 0.7x\text{ = 0.36(10+x) } \\ 0.7x\text{ = 3.6 + 0.36x} \\ 0.7x-0.36x\text{ = 3.6} \\ 0.34x\text{ = 3.6} \\ x\text{ =}(3.6)/(0.34)\text{ =10.58} \\ x\text{ = 10.58} \\ from\text{ equation 1} \\ h\text{ =0.7}x \\ h\text{ = 0.7 x 10.58} \\ h\text{ = 7.406} \end{gathered}$

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