+0

# Triangle $ABC$ is inscribed in equilateral triangle $PQR$, as shown. If $PC = 3$, $BP = CQ = 2$, and $\angle ACB = 90^\circ$, then compute $+2 1000 6 +598 Triangle$ABC$is inscribed in equilateral triangle$PQR$, as shown. If$PC = 3$,$BP = CQ = 2$, and$\angle ACB = 90^\circ$, then compute$AQ$. Dec 5, 2017 ### 6+0 Answers #1 +98196 +3 Using the Law of Cosines BC^2 = BP^2 + PC^2 - 2(BP)(PC)cos (60) BC^2 = 2^2 + 3^2 - 2(2)(3)(1/2) BC^2 = 13 - 6 BC = √7 Using the Law of Sines sin BPC / BC = sin BCP / BP √3/[2√7] = sin BCP / 2 sin BCP = √(3/7) arcsin [ √(3/7) ] = BCP ≈ 40.89° And BCP and ACQ are complementary....so ACQ ≈ 49.106° Which means that CAQ = 180 - 60 - 49.106 ≈ 70.893° So.... using the Law of Sines again AC / sin AQC = CQ / sin CAQ AC = √3/ sin (70.893) ≈ 1.833 And using the Law of Sines once more AC / sin AQC = AQ / sin ( ACQ) AQ = sin (ACQ) * AC / sin (AQC) AQ = sin (49.106) * 1.833 / sin (60) ≈ 1.5996 = 1.6 Dec 5, 2017 #2 +21869 +3 Triangle$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

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

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

Dec 5, 2017
#4
+21869
+2

CPhill, thank you very much

heureka  Dec 5, 2017
#5
+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
+98196
+1

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

Dec 5, 2017