Question

Jane is using a sextant to measure her distance from a building of height 270 feet. Her eyes

are 5 feet 3 inches above the ground. The angle of elevation from her viewpoint using the
sextant is 40°. How far away, measured to the nearest foot, is Jane from the building?
a) 221 feet b) 203 feet c) 170 feet d) 316 feet e) none of these.

Answer 1

Answer: d) 316 feet

Explanation:

A right angled triangle ABC is formed.

BC represents the height of the building minus the height of Jane's eyes from the ground. AB represents Jane's distance from the building.

12 inches = 1 feet

3 inches = 3/12 = 0.25 feet

Therefore, the height of her eyes from the ground is 5 + 0.25 = 5.25 feet

Therefore, BC = 270 - 5.25 = 264.75 feet

To determine AB, we would apply the the tan trigonometric ratio which is expressed as

Tan# = opposite side/adjacent side

Tan 40 = 264.75/AB

AB = 264.75/tan40

AB = 264.75/0.839

AB = 316 feet

Answer 2

d) 316 feet

Explanation:

Now 3 Inches = 0.25 feet

Therefore: 5 feet 3 inches =5.25 feet

The problem forms a right triangle in which the height of the triangle

= 270 -5.25 =264.75 feet.

Using Trigonometric ratios

`\tan 40^\circ=(264.75)/(y) \\$Cross multiply\\y\tan 40^\circ=264.75\\y=264.75 / \tan 40^\circ\\y=315.5$ feet`

Therefore, Jane is approximately 316 feet from the building.