Difference between revisions of "1978 AHSME Problems/Problem 15"
(Created page with "==Solution== Squaring the equation, we get <cmath>\sin ^2 x + \cos ^2 x + 2 \sin x \cos x = \frac{1}{25} =\Rrightarrow 2\sin x \cos x = -\frac{24}{25} \Rrightarrow \sin x \cos...") |
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+ | == Problem 15 == | ||
+ | |||
+ | If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is | ||
+ | |||
+ | <math>\textbf{(A) }-\frac{4}{3}\qquad | ||
+ | \textbf{(B) }-\frac{3}{4}\qquad | ||
+ | \textbf{(C) }\frac{3}{4}\qquad | ||
+ | \textbf{(D) }\frac{4}{3}\qquad\\ | ||
+ | \textbf{(E) }\text{not completely determined by the given information} </math> | ||
+ | |||
==Solution== | ==Solution== | ||
Squaring the equation, we get | Squaring the equation, we get | ||
Line 15: | Line 25: | ||
~JustinLee2017 | ~JustinLee2017 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1978|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:10, 13 February 2021
Problem 15
If and , then is
Solution
Squaring the equation, we get Recall that , so We can now solve for the values of and Since , we have
~JustinLee2017
See Also
1978 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.