Difference between revisions of "1985 AHSME Problems/Problem 28"
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==Problem== | ==Problem== | ||
− | In <math> \triangle ABC </math>, we have <math> \angle C=3\angle A, a=27 | + | In <math> \triangle ABC </math>, we have <math> \angle C=3\angle A, a=27 </math> and <math> c=48 </math>. What is <math> b </math>? |
<asy> | <asy> | ||
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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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<math> \implies 36\sin^3(A)=11\sin(A) </math>. | <math> \implies 36\sin^3(A)=11\sin(A) </math>. | ||
− | Therefore, <math> \sin(A)=0, \frac{\sqrt{11}}{6}, </math> or <math> -\frac{\sqrt{11}}{6} </math>. Notice that we must have <math> 0^\circ<A<45^\circ </math> because otherwise <math> A+3A>180^\circ </math>. We can therefore disregard <math> \sin(A)=0 </math> because then <math> A=0 </math> and also we can disregard <math> \sin(A)=-\frac{\sqrt{11}}{6 | + | Therefore, <math> \sin(A)=0, \frac{\sqrt{11}}{6}, </math> or <math> -\frac{\sqrt{11}}{6} </math>. Notice that we must have <math> 0^\circ<A<45^\circ </math> because otherwise <math> A+3A>180^\circ </math>. We can therefore disregard <math> \sin(A)=0 </math> because then <math> A=0 </math> and also we can disregard <math> \sin(A)=-\frac{\sqrt{11}}{6} </math> because then <math> A </math> would be in the third or fourth quadrants, much greater than the desired range. |
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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 <math>A</math> be equal to <math>x</math> degrees. Then angle <math>C</math> is equal to <math>3x</math> degrees, and angle <math>B</math> is equal to <math>180-4x</math> degrees. Let <math>D</math> be a point on side <math>AB</math> such that <math>\angle ACD</math> is equal to <math>x</math> degrees. Because <math>2x+180-4x+\angle CDB=180</math>, angle <math>CDB</math> is equal to <math>2x</math> degrees. We can now see that triangles <math>CDB</math> and <math>CDA</math> are both isosceles, with <math>CB=DB</math> and <math>AD=AC</math>. From isosceles triangle <math>CDB</math>, we now know that <math>BD = 27</math>, and since <math>AB = c = 48</math>, we know that <math>AD = 21</math>. From isosceles triangle <math>CDA</math>, we now know that <math>CD = 21</math>. Applying [[Stewart's Theorem]] on triangle <math>ABC</math> gives us <math>AC = 35</math>, 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}} |
Revision as of 16:44, 20 July 2018
Contents
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 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 be equal to degrees. Then angle is equal to degrees, and angle is equal to degrees. Let be a point on side such that is equal to degrees. Because , angle is equal to degrees. We can now see that triangles and are both isosceles, with and . From isosceles triangle , we now know that , and since , we know that . From isosceles triangle , we now know that . Applying Stewart's Theorem on triangle gives us , 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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