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Question Number 203321 by SonGoku last updated on 16/Jan/24

$$\mathrm{Help}-\mathrm{me}!$$$$$$$$\mathrm{Observe}\mathrm{points}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{below}\mathrm{and}\mathrm{find}\mathrm{the}\mathrm{widthof}\mathrm{a}\mathrm{lake}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{following}\mathrm{data}:$$$$$$$$\left(\mathrm{AB}\right)\mathrm{m};\hat{\mathrm{C}}=39°{52}^{\prime }{12}^{″}$$$$(\mathrm{BC}-257.5)\mathrm{m};\hat{\mathrm{A}}=97°7^{\prime }{56}^{″}$$$$(\mathrm{CA}-30)\mathrm{m};\hat{\mathrm{B}}=42°{59}^{\prime }{52}^{″}$$$$\mathrm{CA}\mathrm{is}\mathrm{the}\mathrm{width}\mathrm{of}\mathrm{the}\mathrm{lake}$$$$$$$${\bullet }^{\mathrm{C}}$$$$$$$$$$$$$$$$$$$$$$$$$$$${\bullet }_{\mathrm{A}}$$$$$$$${\bullet }_{\mathrm{B}}$$

Answered by esmaeil last updated on 17/Jan/24

$$1^o={60}^{{}^{\prime }}={3600}^{\parallel }\to$$$$\stackrel{\wedge }{C}=39+(\frac{52}{60}+\frac{12}{3600})=39.{87}^o$$$$\stackrel{\wedge }{B}=42+(\frac{59}{60}+\frac{52}{3600})\approx 42.9978$$$$\frac{AB}{sinc}=\frac{AC}{sinB}\to AB=\frac{sin(39.87)}{sin(42.9987)}\times 30$$$$=28.1993$$

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