# Thread: Height of Flagpole

1. ## Height of Flagpole

From a point 100 feet away in front of a public library, the angles of elevation to the base of the flagpole and the top of the flagpole are 28 degrees and 39 degress 45', respectively. The flagpole is mounted on the front of the building's roof. Find height of the flagpole.

Must I change 39 degrees 45' into a degree first?

How is this done?

2. Originally Posted by magentarita
From a point 100 feet away in front of a public library, the angles of elevation to the base of the flagpole and the top of the flagpole are 28 degrees and 39 degress 45', respectively. The flagpole is mounted on the front of the building's roof. Find height of the flagpole.

Must I change 39 degrees 45' into a degree first?

How is this done?
Not necessary . However if u want to do so , note that 1 degree equals 60 mins . So 39 degree 45 ' equals 39.75 degree .

First , find the height of the base of the flagpole from the ground .
tan 28 = $h_1$ / 100

then find the height of the top of the flagpole from the ground
tan39.75 = $h_2$/100

Then $h_2$- $h_1$= height of flagpole
I assume that the height of flagpole you meant is the length of flagpole .

3. Originally Posted by magentarita
From a point 100 feet away in front of a public library, the angles of elevation to the base of the flagpole and the top of the flagpole are 28 degrees and 39 degress 45', respectively. The flagpole is mounted on the front of the building's roof. Find height of the flagpole.

Must I change 39 degrees 45' into a degree first?

How is this done?
1. $45' = \left(\dfrac{45}{60}\right)^\circ = 0.75^\circ$

2. You are dealing with 2 right triangles. Use tan-function to calculate the second leg:

The length of the flagpole is:

$f = 100\cdot \tan(39.75^\circ) - 100\cdot \tan(28^\circ) = 100 \cdot (\tan(39.75^\circ) - \tan(28^\circ)) = 29.998...\approx 30'$

4. ## ok............

Originally Posted by mathaddict
Not necessary . However if u want to do so , note that 1 degree equals 60 mins . So 39 degree 45 ' equals 39.75 degree .

First , find the height of the base of the flagpole from the ground .
tan 28 = $h_1$ / 100

then find the height of the top of the flagpole from the ground
tan39.75 = $h_2$/100

Then $h_2$- $h_1$= height of flagpole
I assume that the height of flagpole you meant is the length of flagpole .
Thanks for your time and effort.

5. ## nice...

Originally Posted by earboth
1. $45' = \left(\dfrac{45}{60}\right)^\circ = 0.75^\circ$

2. You are dealing with 2 right triangles. Use tan-function to calculate the second leg:

The length of the flagpole is:

$f = 100\cdot \tan(39.75^\circ) - 100\cdot \tan(28^\circ) = 100 \cdot (\tan(39.75^\circ) - \tan(28^\circ)) = 29.998...\approx 30'$
Nice picture and reply.

### from a point 100 feet in front of a public library

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