Difference between revisions of "1985 AHSME Problems/Problem 28"
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<math> \mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined} </math> | <math> \mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined} </math> | ||
− | ==Solution== | + | ==Solution 1== |
From the Law of Sines, we have <math> \frac{\sin(A)}{a}=\frac{\sin(C)}{c} </math>, or <math> \frac{\sin(A)}{27}=\frac{\sin(3A)}{48}\implies 9\sin(3A)=16\sin(A) </math>. | From the Law of Sines, we have <math> \frac{\sin(A)}{a}=\frac{\sin(C)}{c} </math>, or <math> \frac{\sin(A)}{27}=\frac{\sin(3A)}{48}\implies 9\sin(3A)=16\sin(A) </math>. | ||
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Therefore, <math> \frac{\frac{\sqrt{11}}{6}}{27}=\frac{\frac{35\sqrt{11}}{162}}{b}\implies b=\frac{6\cdot27\cdot35}{162}=35, \boxed{\text{B}} </math>. | Therefore, <math> \frac{\frac{\sqrt{11}}{6}}{27}=\frac{\frac{35\sqrt{11}}{162}}{b}\implies b=\frac{6\cdot27\cdot35}{162}=35, \boxed{\text{B}} </math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Let angle A be equal to <math>x</math> degrees. Then angle C is equal to <math>3x</math> degrees, and angle B is equal to <math>180-4x</math> degrees. Let D be a point on side AB such that angle CDA is equal to <math>x</math> degrees, and angle CDB is equal to <math>2x</math> degrees. We can now see that triangles CDB and CDA are both isosceles. From isosceles triangle CDB, we now know that BD = 27, and since AB = <math>c</math> = 48, we know that AD = 21. From isosceles triangle CDA, we now know that CD = 21. Applying Stewart's Theorem on triangle ABC gives us AC = 35, which is <math>\boxed{\text{B}}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=27|num-a=29}} | {{AHSME box|year=1985|num-b=27|num-a=29}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:48, 14 February 2014
Contents
[hide]Problem
In , we have and . What is ?
Solution 1
From the Law of Sines, we have , or .
We now need to find an identity relating and . We have
.
Thus we have
.
Therefore, or . Notice that we must have because otherwise . We can therefore disregard because then and also we can disregard $\sin(A)=-\frac{\sqrt{11}}{6}}$ (Error compiling LaTeX. Unknown error_msg) because then would be in the third or fourth quadrants, much greater than the desired range.
Therefore, , and . Going back to the Law of Sines, we have .
We now need to find .
.
Therefore, .
Solution 2
Let angle A be equal to degrees. Then angle C is equal to degrees, and angle B is equal to degrees. Let D be a point on side AB such that angle CDA is equal to degrees, and angle CDB is equal to degrees. We can now see that triangles CDB and CDA are both isosceles. From isosceles triangle CDB, we now know that BD = 27, and since AB = = 48, we know that AD = 21. From isosceles triangle CDA, we now know that CD = 21. Applying Stewart's Theorem on triangle ABC gives us AC = 35, which is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 |
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