3D Geometry
Angles in 3D Space
Grade 12
Question:
<p>In a cubical hall ABCD-PQRS with each side 10 m, G is the centre of the wall BCRQ and T is the mid-point of the side AB. The angle of elevation of G at the point T is</p>
<p>(a) \(\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\)</p>
<p>(b) \(\sin^{-1}\left(\frac{\sqrt{3}}{\sqrt{10}}\right)\)</p>
<p>(c) \(\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)\)</p>
<p>(d) \(\cos^{-1}\left(\frac{\sqrt{5}}{\sqrt{10}}\right)\)</p>
Step-by-Step Solution
Key Concept: Use coordinate geometry or 3D distance formula combined with Pythagoras theorem to find the actual and horizontal distances, then compute angle of elevation.
Solution: Let H be the mid-point of BC. Since \(\angle TBH = 90°\), \((TH)^2 = (BT)^2 + (BH)^2 = 5^2 + 5^2 = 50\) Also, \(\angle THG = 90°\), \((TG)^2 = (TH)^2 + (GH)^2 = 50 + 25 = 75\) Let \(\theta\) be the required angle of elevation of G at T. Then, \(\sin \theta = \frac{GH}{TG} = \frac{5}{\sqrt{75}} = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}}\) ∴ \(\theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Correct Answer: B