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Help-me-Observe-points-A-B-and-C-below-and-find-the-widthof-a-lake-according-to-the-following-data-AB-m-C-39-52-12-BC-257-5-m-A-97-7-56-CA-30-m-B-42-59-52-CA-is-t




Question Number 203321 by SonGoku last updated on 16/Jan/24
Help-me!     Observe points A, B and C below and find the widthof a lake according to the following data:     (AB)m; C^�  = 39°52′12′′  (BC − 257.5)m; A^�  = 97°7′56′′  (CA − 30)m; B^�  = 42°59′52′′  CA is the width of the lake      •^C                     •_A                                         •_B
$$\mathrm{Help}-\mathrm{me}! \\ $$$$\: \\ $$$$\mathrm{Observe}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{below}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{widthof}\:\mathrm{a}\:\mathrm{lake}\:\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{following}\:\mathrm{data}: \\ $$$$\: \\ $$$$\left(\mathrm{AB}\right)\mathrm{m};\:\hat {\mathrm{C}}\:=\:\mathrm{39}°\mathrm{52}'\mathrm{12}'' \\ $$$$\left(\mathrm{BC}\:−\:\mathrm{257}.\mathrm{5}\right)\mathrm{m};\:\hat {\mathrm{A}}\:=\:\mathrm{97}°\mathrm{7}'\mathrm{56}'' \\ $$$$\left(\mathrm{CA}\:−\:\mathrm{30}\right)\mathrm{m};\:\hat {\mathrm{B}}\:=\:\mathrm{42}°\mathrm{59}'\mathrm{52}'' \\ $$$$\mathrm{CA}\:\mathrm{is}\:\mathrm{the}\:\mathrm{width}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lake}\: \\ $$$$\: \\ $$$$\bullet^{\mathrm{C}} \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\bullet_{\mathrm{A}} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bullet_{\mathrm{B}} \\ $$
Answered by esmaeil last updated on 17/Jan/24
1^o =60^′ =3600^∥ →  C^∧ =39+(((52)/(60))+((12)/(3600)))=39.87^o   B^∧ =42+(((59)/(60))+((52)/(3600)))≈42.9978  ((AB)/(sinc))=((AC)/(sinB))→AB=((sin(39.87))/(sin(42.9987)))×30  =28.1993
$$\mathrm{1}^{{o}} =\mathrm{60}^{'} =\mathrm{3600}^{\shortparallel} \rightarrow \\ $$$$\overset{\wedge} {{C}}=\mathrm{39}+\left(\frac{\mathrm{52}}{\mathrm{60}}+\frac{\mathrm{12}}{\mathrm{3600}}\right)=\mathrm{39}.\mathrm{87}^{{o}} \\ $$$$\overset{\wedge} {{B}}=\mathrm{42}+\left(\frac{\mathrm{59}}{\mathrm{60}}+\frac{\mathrm{52}}{\mathrm{3600}}\right)\approx\mathrm{42}.\mathrm{9978} \\ $$$$\frac{{AB}}{{sinc}}=\frac{{AC}}{{sinB}}\rightarrow{AB}=\frac{{sin}\left(\mathrm{39}.\mathrm{87}\right)}{{sin}\left(\mathrm{42}.\mathrm{9987}\right)}×\mathrm{30} \\ $$$$=\mathrm{28}.\mathrm{1993} \\ $$

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