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Difference between revisions of "2024 AMC 12A Problems/Problem 10"

(→‎Solution 10 (don't do this))
m (→‎Solution 2: Scaling and combining triangles)
 
(31 intermediate revisions by 4 users not shown)
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<math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math>
 
<math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math>
 +
 
==Solution 1==
 
==Solution 1==
 
From the question,
 
From the question,
Line 10: Line 11:
 
<cmath>\tan(\alpha+\beta)=\frac{4}{3}</cmath>
 
<cmath>\tan(\alpha+\beta)=\frac{4}{3}</cmath>
 
<cmath>\alpha+\beta=\tan^{-1}\left(\frac{4}{3}\right)</cmath>
 
<cmath>\alpha+\beta=\tan^{-1}\left(\frac{4}{3}\right)</cmath>
 +
<cmath>\alpha+\beta=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right)</cmath>
 
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
 
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
 
<cmath>\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}</cmath>
 
<cmath>\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}</cmath>
Line 15: Line 17:
 
~lptoggled
 
~lptoggled
  
−
==Solution 2: Trial and Error ==  
+
==Solution 2: Scaling and combining triangles==
 +
 
 +
We can scale the <math>3</math>-<math>4</math>-<math>5</math> triangle up by a factor of <math>6</math> to make its side lengths <math>18,24,</math> and <math>30,</math> then glue its side of length <math>24</math> to the corresponding side in the <math>7</math>-<math>24</math>-<math>25</math> triangle:
 +
 
 +
<asy>
 +
pair A = (0,0);
 +
pair B = (18,0);
 +
pair C = (25,0);
 +
pair D = (18,24);
 +
 
 +
draw(A--C--D--cycle);
 +
draw(B--D);
 +
 
 +
draw(rightanglemark(C,B,D,50));
 +
 
 +
label("A", A, SW);
 +
label("B", B, S);
 +
label("C", C, SE);
 +
label("D", D, N);
 +
 
 +
label("18", A--B, S);
 +
label("7", B--C, S);
 +
label("25", C--D, NE);
 +
label("30", D--A, NW);
 +
label("24", B--D, W);
 +
 
 +
label("$\alpha$", D, 6*dir(250));
 +
label("$\beta$", D, 9*dir(278));
 +
label("$\frac{\pi}{2} - \alpha$", A, 4*dir(25));
 +
</asy>
 +
 
 +
Angles <math>\angle DAB</math> and <math>\angle BDA</math> are complementary in <math>\triangle ABD,</math> so <math>\angle DAB = \frac{\pi}{2} - \alpha.</math> We also have <math>AC = 18 + 7 = 25 = CD,</math> so <math>\triangle ACD</math> is isosceles. That means that its base angles <math>\angle CDA</math> and <math>\angle CAD</math> are congruent, so <math>\alpha + \beta = \frac{\pi}{2} - \alpha,</math> and hence <math>\beta = \boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.</math>
 +
 
 +
~MartianTom
 +
 
 +
==Solution 3: Trial and Error ==  
 
Another approach to solving this problem is trial and error, comparing the sine of the answer choices with <math>\sin\beta = \frac{7}{25}</math>. Starting with the easiest sine to compute from the answer choices (option choice D). We get:
 
Another approach to solving this problem is trial and error, comparing the sine of the answer choices with <math>\sin\beta = \frac{7}{25}</math>. Starting with the easiest sine to compute from the answer choices (option choice D). We get:
 
<cmath>\sin{\left(\frac{\alpha}{2}\right)} = \sqrt{\frac{1 - \cos{\alpha}}{2}}</cmath>
 
<cmath>\sin{\left(\frac{\alpha}{2}\right)} = \sqrt{\frac{1 - \cos{\alpha}}{2}}</cmath>
Line 31: Line 68:
 
Since <math>\sin\left(\frac{\pi}{2} - 2\alpha\right)</math> is equal to <math>\sin\beta</math>, option choice C is the correct answer. ~amshah
 
Since <math>\sin\left(\frac{\pi}{2} - 2\alpha\right)</math> is equal to <math>\sin\beta</math>, option choice C is the correct answer. ~amshah
  
−
==Solution 3: ==
+
==Solution 4: ==
  
−
sin(B) = <math>\frac{24}{25}</math> = 2 * <math>\frac{12}{25}</math> = 2 * <math>\frac{3}{5}</math> * <math>\frac{4}{5}</math> = 2 * sin(A) * cos(A) = Sin(2A)  = Cos(90 - 2A)  
+
<math>sin(B) = \frac{24}{25} = 2 \cdot \frac{12}{25} = 2 \cdot \frac{3}{5} \cdot \frac{4}{5}  = 2 \cdot sin(A) \cdot cos(A) = sin(2A)  = cos(90^{\circ} - 2A)</math>
 +
 
 +
Therefore <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
  
−
choice C is the correct answer
 
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
−
==Solution 4: Ptolemy (no trig)==
+
==Solution 5: Ptolemy (no trig)==
 
Let AB have length 15, BC have length 20, AC length 25, AD length 7 and CD length 24. Let x be the measure of segment BD. Thus the measure of angle ACB is <math>\alpha</math> and the measure of angle ACD is <math>\beta</math>. ABCD is a cyclic quadrilateral because angle ABC and angle ADC are right angles. Using Ptolemy's theorem on this quadrilateral yields 25x = 15*24 + 7*20 = 500, or x = 20. This means triangle CBD is isoceles. The perpendicular bisector of CD passes through the center (O) of the circle on which ABCD lies and also passes through B. Let the intersection of the perpendicular bisector of CD and CD be point P. The measure of angle OBC is the same as the measure of the angle OCB which is <math>\alpha</math>, so the measure of angle BOC is <math>{\pi} - {2}{\alpha}</math>, so the measure of angle COP is <math>{2}{\alpha}</math>. Triangle COP is a right triangle with angle OCP being the same as angle ACD (<math>\beta</math>), angle COP being <math>{2}{\alpha}</math>, and angle CPO being <math>\frac{\pi}{2}</math>.  
 
Let AB have length 15, BC have length 20, AC length 25, AD length 7 and CD length 24. Let x be the measure of segment BD. Thus the measure of angle ACB is <math>\alpha</math> and the measure of angle ACD is <math>\beta</math>. ABCD is a cyclic quadrilateral because angle ABC and angle ADC are right angles. Using Ptolemy's theorem on this quadrilateral yields 25x = 15*24 + 7*20 = 500, or x = 20. This means triangle CBD is isoceles. The perpendicular bisector of CD passes through the center (O) of the circle on which ABCD lies and also passes through B. Let the intersection of the perpendicular bisector of CD and CD be point P. The measure of angle OBC is the same as the measure of the angle OCB which is <math>\alpha</math>, so the measure of angle BOC is <math>{\pi} - {2}{\alpha}</math>, so the measure of angle COP is <math>{2}{\alpha}</math>. Triangle COP is a right triangle with angle OCP being the same as angle ACD (<math>\beta</math>), angle COP being <math>{2}{\alpha}</math>, and angle CPO being <math>\frac{\pi}{2}</math>.  
 
So:  
 
So:  
Line 44: Line 82:
 
<cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>~Ilaggo2432
 
<cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>~Ilaggo2432
  
−
==Solution 5(rough value)==
+
==Solution 6(rough value)==
 
given in the question,sin(a)is 3/5 and sin (b) is 7/25,estimate the angle by using the arcsin function. arcsin 3/5 is around 0.643 rad and arcsin 7/25 is around 0.283. Note that pi/2 is closest to 1.57 rad, try one by one and option c suggests that 1.57=2(0.643)+0.283. My first amc 12 edit
 
given in the question,sin(a)is 3/5 and sin (b) is 7/25,estimate the angle by using the arcsin function. arcsin 3/5 is around 0.643 rad and arcsin 7/25 is around 0.283. Note that pi/2 is closest to 1.57 rad, try one by one and option c suggests that 1.57=2(0.643)+0.283. My first amc 12 edit
  
−
==Solution 6 (Angle Addition)==
+
==Solution 7 (Angle Addition)==
 
<math>\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \sin(\alpha+\beta) = \frac{4}{5}</math>. Noticing <math>\frac{4}{5} = \cos(\alpha) = \sin\left(\frac{\pi}{2}-\alpha\right)</math> gives us <math>\alpha + \beta = \frac{\pi}{2} - \alpha</math> so <math>\boxed{\textbf{(C) }\dfrac{\pi}{2} - 2\alpha}</math>
 
<math>\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \sin(\alpha+\beta) = \frac{4}{5}</math>. Noticing <math>\frac{4}{5} = \cos(\alpha) = \sin\left(\frac{\pi}{2}-\alpha\right)</math> gives us <math>\alpha + \beta = \frac{\pi}{2} - \alpha</math> so <math>\boxed{\textbf{(C) }\dfrac{\pi}{2} - 2\alpha}</math>
 
~KEVIN_LIU
 
~KEVIN_LIU
  
−
==Solution 7 (Geometry Solution Without Words)==
+
==Solution 8 (Geometry Solution Without Words)==
 +
 
 +
<cmath>\angle A+ \angle B+ \angle A = \frac{\pi}{2}</cmath>
 +
Therefore <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 
 +
[[Image:2024_amc12A_p10.PNG|thumb|center|400px|]]
 +
 
 +
==Solution 9 (Complex Number)==
 +
 
 +
Define <math>Z_{A} = 4 + 3i = 5e^{iA} </math>
 +
 
 +
and <math>Z_{B} = 24 + 7i = 25e^{iB} </math>
 +
 
 +
Looking at the answer choices, we start by testing out an easier choice, such as C.
 +
 
 +
<math> (4+3i)^2</math>
 +
 
 +
<math>= 16 + 9 \cdot i^2 + 2\cdot12 \cdot i</math>
 +
 
 +
<math>= 16 - 9 + 24i </math>
 +
 
 +
<math>= 7 + 24i </math>
 +
 
 +
Finally:
 +
<math> 25^2 \cdot e^{i(2A+B)} </math>
  
−
<cmath>\angle A+ \angle B+ \angle A = \frac{\pi}{2}</cmath> ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+
<math>= 25e^{i2A} \cdot 25e^{iB} </math>
  
−
[[Image:2024_amc12A_p10.PNG|thumb|center|600px|]]
+
<math>={(5e^{iA})}^2 \cdot 25e^{iB}</math>
  
−
==Solution 8 (Complex Number)==
+
<math>= (7+24i)(24+7i)</math>
  
−
4 + 3i = <math>5e^{iA} </math>
+
<math>=25^2i</math>
  
−
<math> 25e^{i2A} = (4+3i)^2 = 16 + 9 * i^2 + 2*12 * i= 16- 9 + 24i = 7+ 24i </math>
+
<math>=25^2 e^{i\frac{\pi}{2}}</math>
  
−
24 + 7i = <math>25e^{iB} </math>
+
This proves that <math>\angle 2A+ \angle B = \frac{\pi}{2}</math>. (Given that <math>\angle A < \frac{\pi}{2}</math> and <math>\angle B < \frac{\pi}{2}</math>)
−
 
−
<cmath>\angle 2A+ \angle B = \frac{\pi}{2}</cmath>  
 
  
 +
Therefore <cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
−
 
+
==Solution 10 (Double Angle + Co-function Identity)==
−
==Solution 9 (Double Angle + Co-function Identity)==
 
  
 
<cmath>\cos(\alpha) = \frac{4}{5} \quad \text{and} \quad \sin(\alpha) = \frac{3}{5}</cmath>
 
<cmath>\cos(\alpha) = \frac{4}{5} \quad \text{and} \quad \sin(\alpha) = \frac{3}{5}</cmath>
Line 97: Line 158:
 
~sourodeepdeb
 
~sourodeepdeb
  
−
==Solution 10 (don't do this)==
+
==Solution 11 (don't do this)==
−
<math>\sin{x} \approx x-\frac{x^{3}}{6}</math>
+
<cmath>\frac{3}{5} \approx \alpha-\frac{\alpha^{3}}{6}</cmath>
 +
<cmath>-5\alpha^3+30\alpha-18 \approx 0</cmath>
 +
<cmath>\alpha \approx 0.64464</cmath>
 +
<cmath>\frac{7}{25} \approx \beta-\frac{\beta^{3}}{6}</cmath>
 +
<cmath>-25\beta^3+150\beta-42 \approx 0</cmath>
 +
<cmath>\beta \approx 0.28381</cmath>
 +
<cmath>\frac{\pi}{2} - 2 \cdot 0.64464 \approx 0.28151632679 \approx \beta</cmath>
 +
<cmath>\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}</cmath>
 +
 
 +
==Solution 12==
 +
<cmath>\frac{4}{5} \approx 1-\frac{\alpha^2}{2}</cmath>
 +
<cmath>5\alpha^2 \approx 2</cmath>
 +
<cmath>\alpha \approx \sqrt{\frac{2}{5}} \approx 0.63246</cmath>
 +
<cmath>\frac{24}{25} \approx 1-\frac{\beta^2}{2}</cmath>
 +
<cmath>25\beta^2 \approx 2</cmath>
 +
<cmath>\beta \approx \sqrt{\frac{2}{25}} \approx 0.28284</cmath>
 +
<cmath>\frac{\pi}{2} - 2 \cdot 0.63246 \approx 0.30587632679  \approx \beta</cmath>
 +
<cmath>\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}</cmath>
  
−
<math>\frac{3}{5} \approx x-\frac{x^{3}}{6}</math>
 
  
−
<math>-5x^3+30x-18 \approx 0</math>
+
== Video Solution (🚀 2 min solve 🚀) ==
  
−
<math>x \approx 0.64464</math>
+
https://youtu.be/7UPO4bF5VoI
  
−
<math>\pi-2 \cdot 0.28151632679 \approx 0.28151632679 \approx \frac{7}{25}</math>
+
<i>~Education, the Study of Everything</i>
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:42, 6 March 2025

Problem

Let $\alpha$ be the radian measure of the smallest angle in a $3{-}4{-}5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7{-}24{-}25$ right triangle. In terms of $\alpha$, what is $\beta$?

$\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad$

Solution 1

From the question, \[\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}\] \[\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\] \[\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\] \[\tan(\alpha+\beta)=\frac{4}{3}\] \[\alpha+\beta=\tan^{-1}\left(\frac{4}{3}\right)\] \[\alpha+\beta=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right)\] \[\alpha+\beta=\frac{\pi}{2}-\alpha\] \[\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}\]

~lptoggled

Solution 2: Scaling and combining triangles

We can scale the $3$-$4$-$5$ triangle up by a factor of $6$ to make its side lengths $18,24,$ and $30,$ then glue its side of length $24$ to the corresponding side in the $7$-$24$-$25$ triangle:

[asy] pair A = (0,0); pair B = (18,0); pair C = (25,0); pair D = (18,24); draw(A--C--D--cycle); draw(B--D); draw(rightanglemark(C,B,D,50)); label("A", A, SW); label("B", B, S); label("C", C, SE); label("D", D, N); label("18", A--B, S); label("7", B--C, S); label("25", C--D, NE); label("30", D--A, NW); label("24", B--D, W); label("$\alpha$", D, 6*dir(250)); label("$\beta$", D, 9*dir(278)); label("$\frac{\pi}{2} - \alpha$", A, 4*dir(25)); [/asy]

Angles $\angle DAB$ and $\angle BDA$ are complementary in $\triangle ABD,$ so $\angle DAB = \frac{\pi}{2} - \alpha.$ We also have $AC = 18 + 7 = 25 = CD,$ so $\triangle ACD$ is isosceles. That means that its base angles $\angle CDA$ and $\angle CAD$ are congruent, so $\alpha + \beta = \frac{\pi}{2} - \alpha,$ and hence $\beta = \boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.$

~MartianTom

Solution 3: Trial and Error

Another approach to solving this problem is trial and error, comparing the sine of the answer choices with $\sin\beta = \frac{7}{25}$. Starting with the easiest sine to compute from the answer choices (option choice D). We get: \[\sin{\left(\frac{\alpha}{2}\right)} = \sqrt{\frac{1 - \cos{\alpha}}{2}}\] \[= \sqrt{\frac{1 - \frac{4}{5}}{2}}\] \[= \sqrt{\frac{1}{10}}\] \[\neq \frac{7}{25}\]

The next easiest sine to compute is option choice C. \[\sin{\left(\frac{\pi}{2} - 2\alpha\right)} = \sin{\left(\frac{\pi}{2}\right)}\cos{\left(2\alpha\right)}\] \[=\cos{2\alpha}\] \[=\cos^2{\alpha} - \sin^2{\alpha}\] \[=\frac{16}{25} - \frac{9}{25}\] \[=\frac{7}{25}\]

Since $\sin\left(\frac{\pi}{2} - 2\alpha\right)$ is equal to $\sin\beta$, option choice C is the correct answer. ~amshah

Solution 4:

$sin(B) = \frac{24}{25} = 2 \cdot \frac{12}{25} = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = 2 \cdot sin(A) \cdot cos(A) = sin(2A) = cos(90^{\circ} - 2A)$

Therefore \[\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}\]

~luckuso

Solution 5: Ptolemy (no trig)

Let AB have length 15, BC have length 20, AC length 25, AD length 7 and CD length 24. Let x be the measure of segment BD. Thus the measure of angle ACB is $\alpha$ and the measure of angle ACD is $\beta$. ABCD is a cyclic quadrilateral because angle ABC and angle ADC are right angles. Using Ptolemy's theorem on this quadrilateral yields 25x = 15*24 + 7*20 = 500, or x = 20. This means triangle CBD is isoceles. The perpendicular bisector of CD passes through the center (O) of the circle on which ABCD lies and also passes through B. Let the intersection of the perpendicular bisector of CD and CD be point P. The measure of angle OBC is the same as the measure of the angle OCB which is $\alpha$, so the measure of angle BOC is ${\pi} - {2}{\alpha}$, so the measure of angle COP is ${2}{\alpha}$. Triangle COP is a right triangle with angle OCP being the same as angle ACD ($\beta$), angle COP being ${2}{\alpha}$, and angle CPO being $\frac{\pi}{2}$. So: \[\beta + {2}\alpha + \frac{\pi}{2} = \pi\] \[\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}\]~Ilaggo2432

Solution 6(rough value)

given in the question,sin(a)is 3/5 and sin (b) is 7/25,estimate the angle by using the arcsin function. arcsin 3/5 is around 0.643 rad and arcsin 7/25 is around 0.283. Note that pi/2 is closest to 1.57 rad, try one by one and option c suggests that 1.57=2(0.643)+0.283. My first amc 12 edit

Solution 7 (Angle Addition)

$\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \sin(\alpha+\beta) = \frac{4}{5}$. Noticing $\frac{4}{5} = \cos(\alpha) = \sin\left(\frac{\pi}{2}-\alpha\right)$ gives us $\alpha + \beta = \frac{\pi}{2} - \alpha$ so $\boxed{\textbf{(C) }\dfrac{\pi}{2} - 2\alpha}$ ~KEVIN_LIU

Solution 8 (Geometry Solution Without Words)

\[\angle A+ \angle B+ \angle A = \frac{\pi}{2}\] Therefore \[\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}\] ~luckuso

2024 amc12A p10.PNG

Solution 9 (Complex Number)

Define $Z_{A} = 4 + 3i = 5e^{iA}$

and $Z_{B} = 24 + 7i = 25e^{iB}$

Looking at the answer choices, we start by testing out an easier choice, such as C.

$(4+3i)^2$

$= 16 + 9 \cdot i^2 + 2\cdot12 \cdot i$

$= 16 - 9 + 24i$

$= 7 + 24i$

Finally: $25^2 \cdot e^{i(2A+B)}$

$= 25e^{i2A} \cdot 25e^{iB}$

$={(5e^{iA})}^2 \cdot 25e^{iB}$

$= (7+24i)(24+7i)$

$=25^2i$

$=25^2 e^{i\frac{\pi}{2}}$

This proves that $\angle 2A+ \angle B = \frac{\pi}{2}$. (Given that $\angle A < \frac{\pi}{2}$ and $\angle B < \frac{\pi}{2}$)

Therefore \[\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}\]

~luckuso

Solution 10 (Double Angle + Co-function Identity)

\[\cos(\alpha) = \frac{4}{5} \quad \text{and} \quad \sin(\alpha) = \frac{3}{5}\]

Use a double angle identity to calculate $\cos(2\alpha)$:

\[\cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha) = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}\]

We know that:

\[\sin(\beta) = \frac{7}{25}\]

Equate both to each other:

\[\cos(2\alpha) = \sin(\beta)\]

Apply the co-function identity, $\cos(2\alpha) = \sin\left(\frac{\pi}{2} - 2\alpha\right)$:

\[\sin\left(\frac{\pi}{2} - 2\alpha\right) = \sin(\beta)\]

Hence,

\[\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}\]

~sourodeepdeb

Solution 11 (don't do this)

\[\frac{3}{5} \approx \alpha-\frac{\alpha^{3}}{6}\] \[-5\alpha^3+30\alpha-18 \approx 0\] \[\alpha \approx 0.64464\] \[\frac{7}{25} \approx \beta-\frac{\beta^{3}}{6}\] \[-25\beta^3+150\beta-42 \approx 0\] \[\beta \approx 0.28381\] \[\frac{\pi}{2} - 2 \cdot 0.64464 \approx 0.28151632679 \approx \beta\] \[\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}\]

Solution 12

\[\frac{4}{5} \approx 1-\frac{\alpha^2}{2}\] \[5\alpha^2 \approx 2\] \[\alpha \approx \sqrt{\frac{2}{5}} \approx 0.63246\] \[\frac{24}{25} \approx 1-\frac{\beta^2}{2}\] \[25\beta^2 \approx 2\] \[\beta \approx \sqrt{\frac{2}{25}} \approx 0.28284\] \[\frac{\pi}{2} - 2 \cdot 0.63246 \approx 0.30587632679 \approx \beta\] \[\beta=\boxed{(\mathbf{C}) \frac{\pi}{2} - 2\alpha}\]


Video Solution (🚀 2 min solve 🚀)

https://youtu.be/7UPO4bF5VoI

~Education, the Study of Everything

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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