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Question Number 12962 by tawa last updated on 07/May/17

Answered by sandy_suhendra last updated on 09/May/17

Commented by sandy_suhendra last updated on 09/May/17

a) from picture (i) T = top of the flagstaff PR=(√(100^2 +70^2 )) = 122.07 m tan 15° = ((TP)/(PR)) ⇒ TP=PR tan 15° =122.07×0.268=32.7 m b) from picture (ii) MP=(1/2)PQ=50 m tan α=((TP)/(MP))=((32.7)/(50))=0.654 α=33.2°

$$\left.\mathrm{a}\right)\:\mathrm{from}\:\mathrm{picture}\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\mathrm{T}\:=\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{flagstaff} \\ $$$$\:\:\:\:\:\:\mathrm{PR}=\sqrt{\mathrm{100}^{\mathrm{2}} +\mathrm{70}^{\mathrm{2}} \:}\:=\:\mathrm{122}.\mathrm{07}\:\mathrm{m} \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{15}°\:=\:\frac{\mathrm{TP}}{\mathrm{PR}}\:\Rightarrow\:\mathrm{TP}=\mathrm{PR}\:\mathrm{tan}\:\mathrm{15}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{122}.\mathrm{07}×\mathrm{0}.\mathrm{268}=\mathrm{32}.\mathrm{7}\:\mathrm{m}\:\:\:\:\: \\ $$$$\left.\mathrm{b}\right)\:\mathrm{from}\:\mathrm{picture}\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\mathrm{MP}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{PQ}=\mathrm{50}\:\mathrm{m} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:\alpha=\frac{\mathrm{TP}}{\mathrm{MP}}=\frac{\mathrm{32}.\mathrm{7}}{\mathrm{50}}=\mathrm{0}.\mathrm{654} \\ $$$$\:\:\:\:\:\:\:\alpha=\mathrm{33}.\mathrm{2}° \\ $$

Commented by tawa tawa last updated on 10/May/17

wow, God bless you sir.

$$\mathrm{wow},\:\mathrm{G}\boldsymbol{{od}}\:\boldsymbol{{bless}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}}. \\ $$

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