Difference between revisions of "1984 AHSME Problems/Problem 23"

Revision as of 21:54, 8 July 2011

Problem

$\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}$ equals

$\mathrm{(A) \ }\tan{10^\circ}+\tan{20^\circ} \qquad \mathrm{(B) \ }\tan{30^\circ} \qquad \mathrm{(C) \ } \frac{1}{2}(\tan{10^\circ}+\tan{20^\circ}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ}$

Solution

From the sum-to-product formulas, we have $\sin{a}+\sin{b}=2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$ . So, $\sin{10^\circ}+\sin{20^\circ}=2\sin{15^\circ}\cos{5^\circ}$ . Also from the sum-to-product formulas, we have $\cos{a}+\cos{b}=2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$ , so $\cos{10^\circ}+\cos{20^\circ}=2\cos{15^\circ}\cos{5^\circ}$ . Substituting these back into the original fraction, we have $\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}=\frac{2\sin{15^\circ}\cos{5^\circ}}{2\cos{15^\circ}\cos{5^\circ}}=\frac{\sin{15^\circ}}{\cos{15^\circ}}=\tan{15^\circ}, \boxed{\text{D}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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Difference between revisions of "1984 AHSME Problems/Problem 23"

Revision as of 21:54, 8 July 2011

Problem

Solution

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