A trigonometry question

[jolliebollie]

A wire is attached to the top of the tower and to a point on the ground that is 38 m from the base of the tower. If the wire makes a 68° angle with the ground, how tall is the tower?

 

[Deleted member 4993]

A wire is attached to the top of the tower and to a point on the ground that is 38 m from the base of the tower. If the wire makes a 68° angle with the ground, how tall is the tower?

Draw a sketch of a vertical right-angled-triangle with base angles being 90 & 68. The length of the base is 38. Define the height as 'h'.

Now use the definition of 'tan' to calculate 'h'.

Please share your work/thoughts about this problem.

 

[The Highlander]

A wire is attached to the top of the tower and to a point on the ground that is 38 m from the base of the tower. If the wire makes a 68° angle with the ground, how tall is the tower?

Hi jolliebollie,
Please make sure you have studied and learned everything in the factsheet (below). It's a good idea to copy it into (all) your workbooks. ?
Then follow the advice given to you above. ?
(NB: Be careful to note that the Tower Height that Subhotosh has suggested you call "h" is actually named "y" (in the factsheet) and the distance provided would be "x"; the angle is, of course, "θ".)

​

 

[Shiloh]

A wire is attached to the top of the tower and to a point on the ground that is 38 m from the base of the tower. If the wire makes a 68° angle with the ground, how tall is the tower?

So \(\displaystyle tan cot(68)= \frac{38}{h}\) where h is the height of the tower. ................................edited

Can you solve for h?

 

[The Highlander]

So $\displaystyle tan(68)= \frac{38}{h}$ where h is the height of the tower. Can you solve for h?

No, Shiloh, that is wrong!
What you could have written (though you should really have left the OP to figure it out for themself!) is:

"$\displaystyle tan(68°)= \frac{h}{38}$ where h is the height of the tower. Can you solve for h?"
​

(The degrees symbol is important too, otherwise the angle might be assumed to be in radians!)
Please don't offer "help" if you're not sure what you're doing yourself.
Thank you.

 

[Steven G]

... or $\displaystyle \cot(68\degree ) = 38/h$, if you insist on using 38/h ..... Edited

 

[The Highlander]

... or $\displaystyle \cot(68)\degree = 38/h$, if you insist on using 38/h

You get back in your corner and stop muddying the waters even further. ???

 

[Otis]

$\displaystyle \cot(68)\degree$

Is that how you approximate -0.49º ?



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