Question #104463
A group of Boy Scouts on a straight trail headed N30.6° E found the trail led through a briar patch. They chose to walk 156 meters due east along the south edge of the briar patch, then due north along the east edge of the patch back to the trail. How
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Expert's answer
2020-03-04T18:01:14-0500

A group of Boy Scouts on a straight trail headed N30.6° E found the trail led through a briar patch. They chose to walk 156 meters due east along the south edge of the briar patch, then due north along the east edge of the patch back to the trail. How much farther did they walk to avoid walking through the briar patch?


Consider the right triangle OAB.\triangle OAB. Given that θ=30.6°,OB=156 m.\theta=30.6\degree, OB=156\ m.

Find BABA


tan(AOB)=BAOB,where AOB=90°θ\tan(\angle AOB)={BA \over OB}, where\ \angle AOB=90\degree-\theta

Then


BA=OBtan(90°θ)=OBcotθBA=OB\tan (90\degree-\theta)=OB\cot\thetaBA=156cot(30.6°)263.78(m)BA=156\cot(30.6\degree)\approx263.78(m)

Find OAOA


cos(AOB)=OBOA\cos(\angle AOB)={OB \over OA}

Then


OA=OBcos(90°θ)=OBsinθOA={OB \over \cos(90\degree-\theta)}={OB \over \sin\theta}

OB+BAOA=OB+OBcotθOBsinθOB+BA-OA=OB+OB\cot\theta-{OB \over \sin\theta}

OB+BAOA=OB(1+cotθ1sinθ)OB+BA-OA=OB(1+\cot\theta-{1 \over \sin\theta})

OB+BAOA=156(1+cot30.6°1sin30.6°)OB+BA-OA=156\big(1+\cot30.6\degree-{1 \over \sin30.6\degree}\big)\approx113.32(m)\approx113.32(m)

They walked 113.32 m113.32\ m farther to avoid walking through the briar patch.



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