Bearing the Giza

Geometry Level 3

A person on a camel observes the top of the Great Pyramid of Giza with a clinometer and finds the angle of elevation to be 3 0 ∘ 30^\circ . After traveling 54 m \rm 54 \ m towards the pyramid, the angle of elevation increases by 7 ° 7° . Let the height of the camel along with the man sitting on it be 3 m \rm 3 \ m . What is the height of the pyramid to the nearest tens place in meters?

(Please don't use a calculator.)


The answer is 140.

Shubhrajit Sadhukhan by Shubhrajit Sadhukhan , 16, India

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2 solutions

K T
Feb 8, 2021

  • calculated tan 37° (using a 2nd order Taylor expansion in 30°)
  • calculated the original distance d d from d ⋅ tan ⁡ = ( d − 54 ) ⋅ tan ⁡ 37 ° d\cdot \tan=(d-54) \cdot \tan 37°
  • calculated the height as (h=3+d \cdot \tan 30°)
  • got 137m, rounded to 140m
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From the 3-4-5 triangle, you could have got tan ⁡ 37 ° ≈ 3 4 \tan 37° ≈ \dfrac34 .

See

Shubhrajit Sadhukhan - 4 months ago

Let the height of the pyramid be a and the hypotenuse of the triangle with 37° be c.

Using the Difference Angle formula ,

sin ⁡ 7 ° = sin ⁡ ( 37 − 30 ) ° = sin ⁡ 37 ° cos ⁡ 30 ° − sin ⁡ 30 ° cos ⁡ 37 ° = 3 5 . 3 2 − 1 2 . 4 5 = 3 3 − 4 10 \begin{aligned} \sin 7° &= \sin (37 - 30)° \\ & = \sin37° \cos30° - \sin30° \cos37° \\ & = \frac35.\frac{\sqrt3}2 - \frac12.\frac45 \\ & = \frac{3\sqrt3 - 4}{10} \end{aligned}

Using the Sine Rule ,

54 sin ⁡ 7 ° = c sin ⁡ 30 ° c = 270 3 3 − 4 sin ⁡ 37 ° c = 3 5 ( 270 3 3 − 4 ) a ≈ 135 \begin{aligned} \dfrac{54}{\sin7°} &= \dfrac{c}{\sin30°} \\ c &= \dfrac{270}{3\sqrt3 - 4} \\ \sin37° c &= \dfrac35 \left(\dfrac{270}{3\sqrt3 - 4}\right) \\ a &≈ 135 \end{aligned}

So the height is 135 + 3 = 138 m 135 + 3 = 138 \si{m} .

In this way, the height of any structure can be measured with the help of trigonometry!

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