A vertical pole consists of two parts, each one half of the whole pole.

[Stein]

Application of Double Angles (Addition Formulas): A vertical pole consists of two parts, each one half of the whole pole. At a point in the horizontal plane which passes through the foot of the pole and 36 m from it, the upper half of the pole an angle whose tangent is 1/3. How high is the pole?
Answer: 36m

Please help!!! I need a detailed solution. Thanks. Feel free to correct me if the answer is wrong.

 

[Deleted member 4993]

A vertical pole consists of two parts, each one half of the whole pole. At a point in the horizontal plane which passes through the foot of the pole and 36 m from it, the upper half of the pole an angle whose tangent is 1/3. How high is the pole?
Answer: 36m

Please help!!! I need a detailed solution. Thanks. Feel free to correct me if the answer is wrong.

You have posted five problems without showing a line of work!

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[Stein]

My sincerest apology, Mr. Khan. As I said I'm a college freshman and we're currently in the application of double angle identities.

To start off,
I can picture out the drawing.

And then, I was stucked I don't know if I should represent "theta" as the angle in the lower half or something. And If that's the case how can i apply the double angle identities? Please help me. Sorry if its unclear to you or it troubles you to understand what I'm trying to say. Thanks in advance.

 

[Deleted member 4993]

My sincerest apology, Mr. Khan. As I said I'm a college freshman and we're currently in the application of double angle identities.

To start off,
I can picture out the drawing.
View attachment 4406
And then, I was stucked I don't know if I should represent "theta" as the angle in the lower half or something. And If that's the case how can i apply the double angle identities? Please help me. Sorry if its unclear to you or it troubles you to understand what I'm trying to say. Thanks in advance.

Lets call

top of the pole = T
Middle of the pole = M
bottom of the pole = B

The point on the ground = G

Then

mTGB = α → tan(α) = H/6

mMGB = ß → tan(ß) = (H/2)/6

mMGT = α - ß

tan(α - ß) = [tan(α) - tan(ß)]/[1 + tan(α)*tan(ß)] = 1/3

Now continue.....

 

[Stein]

I will substitute the value of tan (a) and tan (b)...

tan(a-b) = tan(a)-tan(b)
-------------
1+tan(a)tan(b)

= (h/6)-(h/12)
-------------
1+(h/6)(h/12)

= (h/12)
-------------
{[(h)^2 +72]/72]}

= 6h 1
------------ = -
[(h)^2 +72] 3

=(h)^2 - 18h + 72 = 0

Am i correct? what will i do then? Thanks mr. khan for your time i really appreciate it.

 

[Deleted member 4993]

I will substitute the value of tan (a) and tan (b)...

tan(a-b) = tan(a)-tan(b)
-------------
1+tan(a)tan(b)

= (h/6)-(h/12)
-------------
1+(h/6)(h/12)

= (h/12)
-------------
{[(h)^2 +72]/72]}

= 6h 1
------------ = -
[(h)^2 +72] 3

=(h)^2 - 18h + 72 = 0

Am i correct? what will i do then? Thanks mr. khan for your time i really appreciate it.

Assuming your calculations are correct, you have a quadratic equation in 'h' → h² - 18h + 72 = 0

Using your favorite method, solve for 'h'.

 

[Stein]

i'll show to you my complete solution...

tan(a-b) = tan(a)-tan(b)/[1+tan(a)*tan(b)]
=[(h/6)-(h/12)]/{1+/([h/6)(h/12)]} i get the LCD
=[h/12]/{[72+(h)^2]/72}
=[6h]/[{72+(h)^2]} = 1/3
i arrived in this equation...
(h)^2 - 18h + 72 =0
can you lend me a hand in solving this quad equation mr.khan? because i used the but the answer is not 36. can you show me how you solve it?

 

[Deleted member 4993]

i'll show to you my complete solution...

tan(a-b) = tan(a)-tan(b)/[1+tan(a)*tan(b)]
=[(h/6)-(h/12)]/{1+/([h/6)(h/12)]} i get the LCD
=[h/12]/{[72+(h)^2]/72}
=[6h]/[{72+(h)^2]} = 1/3
i arrived in this equation...
(h)^2 - 18h + 72 =0
can you lend me a hand in solving this quad equation mr.khan? because i used the View attachment 4408 but the answer is not 36. can you show me how you solve it?

At a point in the horizontal plane which passes through the foot of the pole and 36 m from it,

The base of the triangle is 36 m - instead of 6 m.

 

[Stein]

I'm very sorry Mr. Khan for my carelessness!!! I finally got the correct answer thank you very much!!!