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Question-12962




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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