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1999 AHSME Problems/Problem 27

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Problem

In triangle $ABC$ , $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$ . Then $\angle C$ in degrees is

$\mathrm{(A) \ }30 \qquad \mathrm{(B) \ }60 \qquad \mathrm{(C) \ }90 \qquad \mathrm{(D) \ }120 \qquad \mathrm{(E) \ }150$

Solution

Square the given equations and add (simplifying with the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ ):

$\begin{align*} 9\sin^2 A + 16\cos^2 B + 24 \sin A \cos B & = 36 \\+ 9\cos^2 A + 16\sin^2 B + 24 \sin B \cos A & = 1 \...$

Thus $\frac 12 = \sin A \cos B + \sin B \cos A$ . This is the sine addition identity, so $\frac 12 = \sin (A + B) = \sin (180 - C) = \sin C$ . Thus either $C = 30^{\circ}, 150^{\circ}$ .

If $C = 150$ , then $A + B = 30 \Longrightarrow A,B < 30$ , and $\sin A < \frac 12, \cos A < 1$ . The first equation implies $6 = 3 \sin A + 4\cos B < 3\left(\frac 12\right) + 4(1) = 5.5 < 6$ , which is a contradiction; thus $C = 30 \Longrightarrow \mathrm{(A)}$ .

See also

1999 AHSME (Problems)
Preceded by Problem 26	Followed by Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1999_AHSME_Problems/Problem_27"

Category: Introductory Trigonometry Problems

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