Question

Ben was walking down the street and saw a large bird on top of a flagpole. He has an eye for angles of elevation, and calculated that the angle of elevation from his position was 25°. He quickly ran toward the flagpole 20 feet to get a better view of the bird, which he now sees is a 40 pound hummingbird. At this position, he notes that the angle of elevation is 65°.

a) Draw a diagram to represent the situation (2 marks).
B.How far away was he from the flag pole originally? (3 marks)
c)
How tall is the flagpole? (3 marks)

Answer 1

(a) Refer to the attached image.

(b) 25.6 ft

(c) 12.0 ft

Explanation:

We are presented with a scenario in which Ben observes a bird on a flagpole from two different distances and at two different angles of elevation. We are tasked with (a) creating a diagram that illustrates this situation, (b) determining Ben's original distance from the flagpole, and (c) calculating the height of the flagpole. To achieve this, we will use trigonometric relations, specifically the tangent function, to solve for the distances and height.

`\hrulefill`

(a) Draw a diagram to represent the situation.
`\hrulefill`

Refer to the attached image.


`\hrulefill`

(b) How far away was he from the flagpole originally?
`\hrulefill`

Let's use trigonometry.

Recall,

`\tan(\theta)=\frac{\text{Opposite}}{\text{Adjacent}}`

Using △A₁B₁C₁ we can form an equation:

`\Longrightarrow \tan(\alpha)=(B_1C_1)/(A_1B_1) \\\\\\\\\Longrightarrow \tan(25 \textdegree)=(h)/(y)\\\\\\\\\therefore \boxed{\tan(25 \textdegree)=(h)/(x+20)} \ \dots (1)`

Using △A₂B₂C₂ we can form another equation:

`\Longrightarrow \tan(\beta)=(B_2C_2)/(A_2B_2) \\\\\\\\\therefore \boxed{\tan(65 \textdegree)=(h)/(x)} \ \dots (2)`

We can now form a system of equations:

`\Longrightarrow \left\{\begin{array}{ccc}\tan(25 \textdegree)=(h)/(x+20)&\dots (1)\\\\\tan(65 \textdegree)=(h)/(x)& \dots (2)\end{array}\right`

We need to solve for 'x' then using the relationship 'y = x + 20' we can find 'y'. 'y' is Ben's original distance from the base of the flag pole.

Solve each equation for 'h':

`\Longrightarrow \left\{\begin{array}{ccc}h=(x+20)\tan(25 \textdegree)&\dots (3)\\\\h=x\tan(65 \textdegree)& \dots (4)\end{array}\right`

Now setting the equations equal to each other:

`\Longrightarrow x\tan(65 \textdegree)=(x+20)\tan(25 \textdegree)`

Solving for 'x':

`\Longrightarrow x\tan(65 \textdegree)=(x+20)\tan(25 \textdegree)\\\\\\\\\Longrightarrow x\tan(65 \textdegree)=x\tan(25 \textdegree)+20\tan(25 \textdegree)\\\\\\\\\Longrightarrow x\tan(65 \textdegree)-x\tan(25 \textdegree)=20\tan(25 \textdegree)\\\\\\\\\Longrightarrow x(\tan(65 \textdegree)-\tan(25 \textdegree))=20\tan(25 \textdegree)\\\\\\\\\Longrightarrow x= (20\tan(25 \textdegree))/(\tan(65 \textdegree)-\tan(25 \textdegree)) \\\\\\\\\therefore x \approx 5.6 \ \text{ft}`

Now using the relationship 'y = x + 20'

`\Longrightarrow y = 5.6 + 20\\\\\\\\\therefore \boxed{y=25.6 \ \text{ft}}`

Thus, Ben was originally approximately 25.6 feet away from the flagpole.

`\hrulefill`

(c) How tall is the flagpole?
`\hrulefill`

Using the value of 'x' from the previous part and the tangent relation from the first position:

`\Longrightarrow h=(x+20)\tan(25 \textdegree)\\\\\\\\\Longrightarrow h=(5.6+20)\tan(25 \textdegree)\\\\\\\\\Longrightarrow h=25.6\tan(25 \textdegree)\\\\\\\\\therefore \boxed{h \approx 12.0 \ \text{ft}}`

So, the height of the flagpole from Ben's eye level is approximately 12.0 feet.