Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>A cone of base radius <em>a</em> has its apex at height <em>h</em> above the centre O of the base. If \(OA = OB = AB = a\) (so triangle OAB is equilateral), and considering triangle OBH where \(\tan 30^\circ = \dfrac{h}{a}\), then <em>h</em> equals:</p>
<p>\(h = a\sqrt{3}\)</p>
<p>\(h = \dfrac{a}{\sqrt{3}}\)</p>
<p>\(h = \dfrac{a}{2}\)</p>
<p>\(h = a\)</p>
Step-by-Step Solution
Key Concept: Recognize that the equilateral triangle OAB with side length a determines the semi-vertical angle of the cone as 30°, and use the relationship tan(semi-vertical angle) = h/a to find h = a/√3.
<p><strong>Step 1:</strong> Understand the geometry. We have an equilateral triangle OAB with OA = OB = AB = a, where O is the center of the base and A, B are points on the base circle (radius a).</p><p><strong>Step 2:</strong> In triangle OBH, where H is directly above O at height h, angle BOH is the semi-vertical angle of the cone. Since OAB is equilateral with OA = OB = a and both lie in the base plane, angle AOB = 60°, making the semi-vertical angle ∠BOH = 30°.</p><p><strong>Step 3:</strong> In right triangle OBH: tan(30°) = h/OB = h/a</p><p><strong>Step 4:</strong> Since tan(30°) = 1/√3, we get: h/a = 1/√3</p><p><strong>Step 5:</strong> Therefore: h = a/√3 = a√3/3</p><p>∴ Answer: B</p>
Correct Answer: B