Trigonometry & Inverse Trigonometry
Angles of Elevation and Depression
Grade 11
Question:
<p>The angle of elevation of tower from a point A due south of it is 30° and from a point B due west of it is 45°. If the height of the tower be 100 m, then AB =</p>
<p>(a) 150 m</p>
<p>(b) 200 m</p>
<p>(c) 173.2 m</p>
<p>(d) 141.4 m</p>
Step-by-Step Solution
Key Concept: Use cotangent of angle of elevation to find horizontal distances, then apply Pythagoras theorem since the points are perpendicular.
<p><strong>Solution:</strong></p><p>Let O be the base of the tower with height h = 100 m.</p><p>From point B due west: $OB = h \cot 45° = 100 \times 1 = 100$ m</p><p>From point A due south: $OA = h \cot 30° = 100 \times \sqrt{3} = 100\sqrt{3}$ m</p><p>Since A is due south and B is due west, OA and OB are perpendicular.</p><p>By Pythagoras theorem:</p><p>$AB = \sqrt{OA^2 + OB^2} = \sqrt{(100\sqrt{3})^2 + 100^2}$</p><p>$= \sqrt{30000 + 10000} = \sqrt{40000} = 200$ m</p><p>∴ Answer is (b) 200 m</p>
Correct Answer: b