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Triangle $ABC$ is inscribed in equilateral triangle $PQR$, as shown. If $PC = 3$, $BP = CQ = 2$, and $\angle ACB = 90^\circ$, then compute $AQ$.

 

 Dec 5, 2017
 #1
avatar+98196 
+3

Using the Law of Cosines

 

BC^2 = BP^2  + PC^2  - 2(BP)(PC)cos (60)

BC^2  =  2^2 + 3^2  - 2(2)(3)(1/2)

BC^2 = 13  - 6

BC  = √7

 

Using the Law of Sines

sin BPC / BC  =  sin BCP / BP

√3/[2√7]  =  sin BCP / 2

sin BCP  = √(3/7)

arcsin [ √(3/7)  ]  = BCP ≈ 40.89°

 

And BCP  and  ACQ  are complementary....so ACQ ≈ 49.106°

Which means that CAQ  =  180 -  60  - 49.106  ≈ 70.893°

 

So.... using the Law of Sines again

 

AC / sin AQC   = CQ / sin CAQ

AC =  √3/ sin (70.893) ≈ 1.833

 

And using the Law of Sines once more

AC / sin AQC  = AQ / sin ( ACQ)

AQ  =  sin (ACQ) * AC / sin (AQC)

AQ =  sin (49.106) * 1.833 / sin (60)    ≈ 1.5996  =   1.6

 

 

cool cool cool

 Dec 5, 2017
 #2
avatar+21869 
+3

Triangle $ABC$ is inscribed in equilateral triangle $PQR$, as shown.

If $PC = 3$, $BP = CQ = 2$, and $\angle ACB = 90^\circ$, then compute $AQ$.

 

Let u = BC

Let v = CA

Let w = AB

Let x = AQ

 

Let PQ = PR = RQ = 5

 

Let BR = PR - BP \(= 5-2=3\)

Let AR = RQ - AQ \(= 5 - x\)

 

Using the Law of Cosines:

\(\begin{array}{|rcll|} \hline u^2 &=& BP^2 + PC^2 - 2\cdot BP \cdot PC\cdot \cos(60^{\circ}) \quad & | \quad 2\cdot \cos(60^{\circ}) = 2\cdot \frac12 = 1 \\ u^2 &=& BP^2 + PC^2 - BP \cdot PC \quad & | \quad BP =2 \quad PC = 3 \\ u^2 &=& 2^2 + 3^2 - 2 \cdot 3 \\ \mathbf{u^2} & \mathbf{=} & \mathbf{2^2 + 3^2 - 6} \\ \hline \end{array}\)

 

Using the Law of Cosines again:

\(\begin{array}{|rcll|} \hline v^2 &=& CQ^2 + AQ^2 - 2\cdot CQ \cdot AQ\cdot \cos(60^{\circ}) \quad & | \quad 2\cdot \cos(60^{\circ}) = 2\cdot \frac12 = 1 \\ v^2 &=& CQ^2 + AQ^2 - CQ \cdot AQ \quad & | \quad CQ =2 \quad AQ = x \\ \mathbf{v^2} & \mathbf{=} & \mathbf{2^2 + x^2 - 2x} \\ \hline \end{array}\)

 

Using the Law of Cosines again:

\(\begin{array}{|rcll|} \hline w^2 &=& BR^2 + AR^2 - 2\cdot BR \cdot AR\cdot \cos(60^{\circ}) \quad & | \quad 2\cdot \cos(60^{\circ}) = 2\cdot \frac12 = 1 \\ w^2 &=& BR^2 + AR^2 - BR \cdot AR \quad & | \quad BR =3 \quad AR = 5-x \\ \mathbf{w^2} & \mathbf{=} & \mathbf{3^2 + (5-x)^2 - 3 \cdot (5-x)} \\ \hline \end{array}\)

 

Using Pythagoras' theorem:

\(\begin{array}{|rcll|} \hline u^2+v^2&=& w^2 \\ (2^2 + 3^2 - 6) + (2^2 + x^2 - 2x) &=& 3^2 + (5-x)^2 - 3 \cdot (5-x) \\ 2^2 + \not{3^2} - 6 + 2^2 + x^2 - 2x &=& \not{3^2} + (5-x)^2 - 3 \cdot (5-x) \\ 2^2 - 6 + 2^2 + x^2 - 2x &=& (5-x)^2 - 3 \cdot (5-x) \\ 2 + x^2 - 2x &=& (5-x)^2 - 3 \cdot (5-x) \\ 2 + \not{x^2} - 2x &=& 25 - 10x + \not{x^2} -15+3x \\ 2 - 2x &=& 25 - 10x -15+3x \\ 2 - 2x &=& 10 - 7x \\ 5x &=& 8 \\ x &=& \frac85 \\ \mathbf{x} & \mathbf{=} & \mathbf{1.6} \\ \hline \end{array}\)

 

AQ is 1.6

 

laugh

 Dec 5, 2017
edited by heureka  Dec 5, 2017
 #3
avatar+98196 
+3

Yep, Heureka....your approach is much better  !!!!

 

 

cool cool cool

 Dec 5, 2017
 #4
avatar+21869 
+2

CPhill, thank you very much

 

laugh

heureka  Dec 5, 2017
 #5
avatar
+1

Here's an alternative, vector type solution.

 

Let \(\displaystyle \hat{\imath}\) be a unit vector in the direction PQ and \(\displaystyle \hat{\jmath}\) a unit vector perpendicular (upwards in the diagram), to this.

 

Then, \(\displaystyle \underline{PB}=2\cos(60)\hat{\imath}+2\sin(60)\hat{\jmath}=\hat{\imath}+\sqrt{3}\hat{\jmath}\) , and \(\displaystyle \underline{PC}=3\hat{\imath}\) . (All angles in degrees.)

 

In the triangle PBC, \(\displaystyle \underline{PB}+\underline{BC}=\underline{PC}\), so \(\displaystyle \underline{BC}=3\hat{\imath}-(\hat{\imath}+\sqrt{3}\hat{\jmath})=2\hat{\imath}-\sqrt{3}\hat{\jmath}\) .

 

Let the vector \(\displaystyle \underline{CA} = \alpha\hat{\imath}+\beta\hat{\jmath}\) , then since CA is at right angles to BC, the scalar product \(\displaystyle \underline{CA}.\underline{BC}=0\) ,

so,  \(\displaystyle 2\alpha-\sqrt{3}\beta = 0\) ...................... (1).

 

The vector \(\displaystyle \underline{AQ} \) will be some fraction of the vector \(\displaystyle \underline{RQ}\) , \(\displaystyle \underline{AQ} = k\underline{RQ}\), say,

so \(\displaystyle \underline{AQ}=k(5\cos(60)\hat{\imath}-5\sin(60)\hat{\jmath})=k \left(\frac{5}{2}\hat{\imath}-\frac{5\sqrt{3}}{2}\hat{\jmath}\right)\).

 

Finally, (almost), in the triangle CAQ,  \(\displaystyle \underline{CA} +\underline{AQ}= \underline{CQ}\) ,

so, \(\displaystyle \alpha\hat{\imath}+\beta\hat{\jmath}+k\left(\frac{5}{2}\hat{\imath}-\frac{5\sqrt{3}}{2}\hat{\jmath}\right)=2\hat{\imath}\) .

 

Equating coefficients, \(\displaystyle \alpha +\frac{5k}{2}=2\quad \text{ so, }\quad\alpha = 2-\frac{5k}{2}\),

and \(\displaystyle \beta -\frac{5k\sqrt{3}}{2}=0 \quad \text{so}\quad \beta = \frac{5k\sqrt{3}}{2}\).

 

Substituting into (1),  \(\displaystyle 2\left(2-\frac{5k}{2}\right)-\sqrt{3}\left(\frac{5k\sqrt{3}}{2}\right)=0\) , so \(\displaystyle 4-\frac{25k}{2}=0,\quad k=\frac{8}{25},\)

so \(\displaystyle AQ = \frac{8}{25}\times5=\frac{8}{5}=1.6\) .

 

Tiggsy

 Dec 5, 2017
 #6
avatar+98196 
+1

Thanks, Tiggsy...that's one I'm going to study closely.....!!!!

 

 

cool cool cool

 Dec 5, 2017

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