Question

A forest ranger in the west observation tower spots a fire 41° north of east. Fifteen miles directly east, the forest ranger in the east tower spots the same fire at 56° north of west. How far away is the ranger who is closest to the fire? Approximate the distance by rounding to the nearest hundredth of a mile. 9.91 mi 11.87 mi 12.53 mi 18.95 mi

Answer 1

9.915 mi and that's all and that helped friend me please thanks stay pretty

Answer 2

9.91 miles

Explanation:

Refer the attached figure

Forest Ranger at point A observes the fire at angle of 41° north of east i.e.∠CAB = 41°

The distance between the two rangers is 15 miles i.e. AB = 15 miles

Forest Ranger at point B observes the fire at at 56° north of west. i.e.∠CBA= 56°

Now we are supposed to find who is closest to the fire

So, we are supposed to find the length of AC and BC

So, first calculate ∠ACB

We will use angle sum property of triangle

Angle sum property of triangle : Sum of all angles of triangle is 180°

So, ∠CBA+∠ACB+∠CAB =180°

56°+∠ACB+41° =180°

97°+∠ACB =180°

∠ACB =180°-97°

∠ACB =83°

Now to find AC and BC we will use law of sines

`(a)/(sin A)=(b)/(Sin B)=(c)/(SinC)`

Refer the attached figure

`(AC)/(sin 56)=(BC)/(Sin 41)=(15)/(Sin83)`

So,
`(BC)/(Sin 41)=(15)/(Sin83)`

`BC=(15)/(Sin83) * Sin 41`

`BC=9.91478`

`(AC)/(sin 56)=(15)/(Sin83)`

`AC=(15)/(Sin83) * sin 56`

`AC=12.5289`

So, BC< AC

So, the ranger who is closest to fire is at a distance of 9.91 miles .

So, Option 1 is true