Answer:
3.4°
Step-by-step explanation:
A 6.0 cm high elevation block is placed under one end of a 1.0 m long track.
Block is 6 cm high. It would stand vertically.
Stand one end of horizontal track 1 m long.
Please see the attachment for graph.
Let the elevation of block is Ф
In ΔABC, ∠B=90°
AB= 6 cm
BC = 1 m
First we change meter to centimeter. 1 m = 100 cm
So, BC = 100 cm
\(\tan\theta=\dfrac{AB}{BC}\)
\(\tan\theta=\dfrac{6}{100}\)
\(\theta=\tan^{-1}(0.06)\)
\(\theta\approx 3.4^\circ\)
Hence, The elevation of the track is 3.4°
