Agroup 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?

  about 113 meters

Step-by-step explanation:

If the straight-line distance is 1, then the distance around is ...

  sin(30.6°) +cos(30.6°) ≈ 1.36978.

The 156 meters due east is the product of the straight-line distance and sin(30.6°), so we have ...

  excess distance = (1.36978 -1)·(straight-line distance)

  = 0.36978·(156 m)/sin(30.6°) = 113.3 m

The group walked about 113 meters farther to avoid the briar patch.

More detail

The sine of an acute angle in a right triangle is the ratio of the opposite side to the hypotenuse. Here, that is ...

  sin(30.6°) = (156 m)/(straight-line distance)

Multiplying by the denominator, and dividing by sin(30.6°), we get ...

  (straight-line distance)·sin(30.6°) = 156 m

  straight-line distance = (156 m)/sin(30.6°) . . . . . . the value used above

actually your concept is wrong

step-by-step explanation:

150% of x =300

150x=300×100

x=(300×100)÷150

x=200