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# Can somebody help me with this Trigonometry question

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From a point A, Tommy notices that the angle of elevation of the top of a building is 20o. He moves 600 meters closer to the building and now measures the angle of elevation to be 36o. How tall is the building?

Guest May 13, 2017

#1
+7336
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Here, CD is the height of the building.

tan 36º = CD / BC

BC tan 36º = CD

BC = CD / tan 36º

tan 20º = CD / (600 + BC)

(600 + BC) tan 20º = CD

BC = CD/tan 20º - 600

$$\frac{CD}{\tan36}=\frac{CD}{\tan20}-600 \\~\\ \frac{CD}{\tan36}-\frac{CD}{\tan20}=-600 \\~\\ CD(\frac1{\tan36}-\frac1{\tan20})=-600 \\~\\ CD=-600\div(\frac1{\tan36}-\frac1{\tan20}) \\~\\ CD \approx 437.606 \text{ meters}$$

hectictar  May 13, 2017
#1
+7336
+4

Here, CD is the height of the building.

tan 36º = CD / BC

BC tan 36º = CD

BC = CD / tan 36º

tan 20º = CD / (600 + BC)

(600 + BC) tan 20º = CD

BC = CD/tan 20º - 600

$$\frac{CD}{\tan36}=\frac{CD}{\tan20}-600 \\~\\ \frac{CD}{\tan36}-\frac{CD}{\tan20}=-600 \\~\\ CD(\frac1{\tan36}-\frac1{\tan20})=-600 \\~\\ CD=-600\div(\frac1{\tan36}-\frac1{\tan20}) \\~\\ CD \approx 437.606 \text{ meters}$$

hectictar  May 13, 2017
#2
+1

Thank you very much!

Guest May 13, 2017