Answer to Question #104463 in Trigonometry for Antoni Perez

Question

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

Answer 1

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 $\triangle OAB.$ Given that $\theta=30.6\degree, OB=156\ m.$

Find $BA$

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

Then

$$BA=OB\tan (90\degree-\theta)=OB\cot\theta$$ $$BA=156\cot(30.6\degree)\approx263.78(m)$$

Find $OA$

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

Then

$$OA={OB \over \cos(90\degree-\theta)}={OB \over \sin\theta}$$

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

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

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

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