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A and B are two point 367m in apart on the same bank of a river. The bearing of a lighthouse observed from A and B are $36^\circ25'$ and $319^\circ25'$ respectively. Find...
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${AD=X\\ BD=367-X\\ \angle A= 93^\circ25’- 36^\circ25’= 57^\circ\\ \angle B= 319^\circ25’ –(180^\circ +93^\circ25’) = 46^\circ}$

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$\text{In } \Delta ADC,\ \tan 57^\circ=\frac{CD}{AD}$ $$X\tan57^\circ=CD \tag{1} {\ \ \ .......(AD=X)}$$

$\text{In } \Delta BDC,\ \tan 46^\circ=\frac{CD}{BD}$ $$(367-X)\tan46^\circ=CD \tag{2} {\ \ \ .......(BD=367-X)}$$

$eq1=eq2\\ X\tan57^\circ=(367-X)\tan46^\circ\\ X=147.56\ m\\ \text{Put X=147.56 in eq1}\\ $

$147\tan 57^\circ=CD\\ $

$CD=227.53\ m$

Width of the river CD = 227.53 m

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