Trigonometry & Inverse Trigonometry
Triangle angle determination
Grade 11
Question:
<p>The sides of a triangle are $\sin a$, $\cos a$, $\sqrt{1 + \sin a \cos a}$ for some $0 < a < \frac{\pi}{2}$ then the greatest angle of the triangle is:</p>
<p>(a) $\frac{\pi}{3}$</p>
<p>(b) $\frac{\pi}{2}$</p>
<p>(c) $\frac{2\pi}{3}$</p>
<p>(d) $\frac{5\pi}{6}$</p>
Step-by-Step Solution
Key Concept: Apply the law of cosines to identify which angle is greatest by comparing the given side lengths.
<p>Let the sides be $a = \sin a$, $b = \cos a$, $c = \sqrt{1 + \sin a \cos a}$.</p><p>The greatest angle is opposite the longest side. Using the law of cosines to find the angle opposite side $c$:</p><p>$c^2 = a^2 + b^2 - 2ab\cos C$</p><p>$1 + \sin a \cos a = \sin^2 a + \cos^2 a - 2\sin a \cos a \cos C$</p><p>$1 + \sin a \cos a = 1 - 2\sin a \cos a \cos C$</p><p>$\cos C = -\frac{1}{2}$, so $C = \frac{2\pi}{3}$</p>
Correct Answer: C