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a person stands at a window on the 9th floor of an office building. he measures the angle of elevation to be 25° and the angle of depression to be 36° of the top and the base of the tower. the person knows that he made the measurements 40 m above the ground.

Determine the height of the tower to the nearest tenth of meter

Please do include the solution, i really need it... please..

Note: Nonesense, Unrespectful and Incorrect Answers will be reported​​


A Person Stands At A Window On The 9th Floor Of An Office Building He Measures The Angle Of Elevation To Be 25 And The Angle Of Depression To Be 36 Of The Top A class=

Sagot :

[tex] \large\underline \mathcal{{QUESTION:}}[/tex]

a person stands at a window on the 9th floor of an office building. he measures the angle of elevation to be 25° and the angle of depression to be 36° of the top and the base of the tower. the person knows that he made the measurements 40 m above the ground.

[tex]\\[/tex]

[tex] \large\underline \mathcal{{SOLUTION:}}[/tex]

First , let's solve for x:

[tex] \sf tan\theta=\frac{40}{x} \: \: \\ \\\sf tan36°=\frac{40}{x} \\ \\ \sf x=\frac{40}{tan36°} \\ \\\sf x=55.055 \\ \\\sf x=55 \: \: \: [/tex]

[tex]\\[/tex]

Then , Solving for y:

[tex] \sf tan\theta=\frac{y}{55} \: \: \: \: \: \: \\ \\ \sf tan25°=\frac{y}{55} \: \: \: \: \\ \\ \sf tan25°(55) = y \\ \\ \sf 25.6469= y \: \: \: \: \: \: \: \: \: \\ \\ \sf 25.65 = y \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

[tex]\\[/tex]

Adding the value two values to get the height:

[tex] \sf Height = 40m+y \: \: \: \: \: \: \: \: \: \: \: \\ \sf Height = 40m+25.65m \\\boxed{ \sf Height = \blue{65.65m }}\: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

[tex]\\[/tex]

[tex] \large\underline \mathcal{{ANSWER:}}[/tex]

  • The height is 65.65 meters
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