SOLUTION: The angular elevation of a tower CD at a place A due south of it is 60°; and at a place B due west of A, the elevation is 30°. If AB = 3 km, what is the height of the tower?

Question 1186123: The angular elevation of a tower CD at a place A due south of it is 60°; and at a place B due west of A, the elevation is 30°. If AB = 3 km, what is the height of the tower?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52777) (Show Source): You can put this solution on YOUR website!

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Let points A,B,C be placed on the ground.
A and B are the observation points.
C is the base of the tower.
D is the top of the tower.

x = distance from A to C
y = distance from B to C
h = height of the tower = distance from C to D
Each distance is in kilometers and each is positive.

Focus on right triangle CAD.
tan(angle) = opposite/adjacent
tan(angle CAD) = CD/CA
tan(60) = h/x
h = x*tan(60)
h = x*sqrt(3)

Now focus on right triangle CBD.
tan(angle) = opposite/adjacent
tan(angle CBD) = CD/CB
tan(30) = h/y
h = y*tan(30)
x*sqrt(3) = y*tan(30) ........... plugged in h = x*sqrt(3)
x*sqrt(3) = y*(1/sqrt(3))
y = x*sqrt(3)*sqrt(3)
y = 3x

Focus on right triangle ABC that is entirely on the ground.
The horizontal leg (that runs east/west) is AB = 3 km.
The leg running north/south is AC = x and the hypotenuse is BC = y = 3x

Use the Pythagorean Theorem for right triangle ABC.
(AB)^2 + (AC)^2 = (BC)^2
(3)^2 + (x)^2 = (3x)^2
9 + x^2 = 9x^2
9x^2-x^2 = 9
8x^2 = 9
x^2 = 9/8
x = sqrt(9/8)
Since x > 0 we ignore the minus of the plus/minus.

Then,
h = x*sqrt(3)
h = sqrt(9/8)*sqrt(3)
h = sqrt((9/8)*3)
h = sqrt(27/8)
h = sqrt(27)/sqrt(8)
h = (3*sqrt(3))/(2*sqrt(2))
h = (3*sqrt(3)*sqrt(2))/(2*sqrt(2)*sqrt(2)) .... rationalizing denominator
h = (3*sqrt(3*2))/(2*2)
h = (3/4)*sqrt(6) km is the exact answer
h = 1.837117 km approximately

The approximate answer varies depending how you round.

------------------------------------------------------------------------------

Another way to solve:

Focus on triangle CAD
tan(60) = h/x
x = h/tan(60)
x = h/sqrt(3)

Focus on triangle CBD
tan(30) = h/y
y = h/tan(30)
y = h*sqrt(3)

The last triangle to focus on is ABC.
It has legs AB = 3 and AC = x. The hypotenuse is BC = y.
Use the Pythagorean Theorem
(AB)^2 + (AC)^2 = (BC)^2
3^2 + x^2 = y^2
3^2 + (h/sqrt(3))^2 = (h*sqrt(3))^2
9 + (h^2)/3 = h^2*3
27 + h^2 = 9h^2
9h^2-h^2 = 27
8h^2 = 27
h^2 = 27/8
h = sqrt(27/8)

The steps from here are the same as the previous section.
You should arrive at
h = (3/4)*sqrt(6) km = 1.837117 km approximately

As of 2025, the world's tallest building is the Burj Khalifa at around 828 meters tall.
The answer we got is over 1000 meters taller, so tower CD is definitely breaking records.
I don't know how realistic it is for a tower to be this tall.

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