+0  
 
+2
560
6
avatar+598 

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

 

michaelcai  Dec 5, 2017
 #1
avatar+87657 
+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

CPhill  Dec 5, 2017
 #2
avatar+19853 
+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

heureka  Dec 5, 2017
edited by heureka  Dec 5, 2017
 #3
avatar+87657 
+3

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

 

 

cool cool cool

CPhill  Dec 5, 2017
 #4
avatar+19853 
+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

Guest Dec 5, 2017
 #6
avatar+87657 
+1

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

 

 

cool cool cool

CPhill  Dec 5, 2017

12 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.