A triangle $$ K L M $$ is similar to triangle $$ P Q R \cdot K L=5 \mathrm{~cm}, L M=40 \mathrm{cn} $$ $$ \angle K L M=46^{\circ} $$. Castruct triangle $$ P Q R $$ sach that $$ Q R=6 \mathrm{~cm} $$

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1. Establish the correspondence between the triangles. Assume that  
  $$
  K \leftrightarrow P,\quad L \leftrightarrow Q,\quad M \leftrightarrow R.
  $$  
 Then side $$KL$$ corresponds to side $$PQ$$ and side $$LM$$ corresponds to side $$QR$$.

2. Given that  
  $$
  KL = 5\,\text{cm},\quad LM = 40\,\text{cm},\quad \angle KLM = 46^\circ,
  $$  
 and that in triangle $$PQR$$ we have  
  $$
  QR = 6\,\text{cm},
  $$  
 determine the similarity (scale) factor from triangle $$KLM$$ to triangle $$PQR$$.

3. Since $$LM \leftrightarrow QR$$, the scale factor is  
  $$
  k = \frac{QR}{LM} = \frac{6}{40} = \frac{3}{20} = 0.15.
  $$

4. Determine the other corresponding sides in triangle $$PQR$$:

a. The side corresponding to $$KL$$ is  
  $$
  PQ = k \cdot KL = 0.15 \times 5 = 0.75\,\text{cm}.
  $$

b. The third side of triangle $$KLM$$ (side $$KM$$) can be found using the Law of Cosines. In triangle $$KLM$$:  
  $$
  KM^2 = KL^2 + LM^2 - 2\cdot KL\cdot LM\cos \angle KLM.
  $$
 Substitute the values:  
  $$
  KM^2 = 5^2 + 40^2 - 2 \cdot 5 \cdot 40\cos 46^\circ = 25 + 1600 - 400\cos 46^\circ.
  $$
 Thus,  
  $$
  KM = \sqrt{1625 - 400\cos 46^\circ}\,.
  ]

c. The side corresponding to $$KM$$ in triangle $$PQR$$ is  
  \[
  PR = k\cdot KM = 0.15\sqrt{1625 - 400\cos 46^\circ}\,.
  ]
 An approximate numerical evaluation gives:
  - First, evaluate $$400\cos46^\circ$$. Using $$\cos46^\circ\approx 0.6947$$:  
   \[
   400\cos46^\circ\approx 277.88\,.
   $$
  - Then,  
   $$
   KM \approx \sqrt{1625 - 277.88} = \sqrt{1347.12} \approx 36.73\,\text{cm}\,.
   $$
  - Finally,  
   $$
   PR \approx 0.15 \times 36.73 \approx 5.51\,\text{cm}\,.
   $$

5. Summary of the sides in triangle $$PQR$$:

- $$PQ \approx 0.75\,\text{cm}$$  
  - $$QR = 6\,\text{cm}$$  
  - $$PR \approx 5.51\,\text{cm}$$

and the corresponding angle $$\angle PQR = 46^\circ$$.

6. Construction Steps for Triangle $$PQR$$:

a. Draw the base $$QR = 6\,\text{cm}$$.

b. At point $$Q$$, construct an angle of $$46^\circ$$.

c. On the ray forming the $$46^\circ$$ angle from $$Q$$, mark point $$P$$ such that $$PQ = 0.75\,\text{cm}$$.

d. With a compass, using point $$P$$ as center, mark an arc with radius $$PR \approx 5.51\,\text{cm}$$.

e. The intersection of this arc with the ray from $$Q$$ (or with an arc centered at $$R$$ if drawn from $$R$$ with radius $$PR$$) locates point $$P$$ precisely completing the triangle.

Thus, triangle $$PQR$$ constructed with $$QR=6\,\text{cm}$$, the angles and proportional sides corresponding to triangle $$KLM$$ is obtained.
To construct triangle $$ PQR $$ similar to triangle $$ KLM $$: 1. **Determine the Scale Factor:** - Given $$ KL = 5\,\text{cm} $$ and $$ QR = 6\,\text{cm} $$, the scale factor $$ k = \frac{6}{5} = 1.2 $$. 2. **Find Corresponding Sides:** - $$ PQ = k \times KL = 1.2 \times 5 = 6\,\text{cm} $$. - $$ PR = k \times KM $$, where $$ KM $$ is calculated using the Law of Cosines: $$ KM = \sqrt{5^2 + 40^2 - 2 \times 5 \times 40 \times \cos46^\circ} \approx 36.73\,\text{cm} $$ Thus, $$ PR = 1.2 \times 36.73 \approx 44.08\,\text{cm} $$. 3. **Construct Triangle $$ PQR $$:** - Draw base $$ QR = 6\,\text{cm} $$. - At point $$ Q $$, construct a $$ 46^\circ $$ angle. - From point $$ Q $$, measure $$ PQ = 6\,\text{cm} $$ along the angle. - From point $$ P $$, draw an arc with radius $$ PR \approx 44.08\,\text{cm} $$ to locate point $$ R $$. 4. **Final Triangle:** - Connect points $$ P $$, $$ Q $$, and $$ R $$ to form triangle $$ PQR $$ with sides $$ PQ = 6\,\text{cm} $$, $$ QR = 6\,\text{cm} $$, and $$ PR \approx 44.08\,\text{cm} $$, and angle $$ \angle PQR = 46^\circ $$.

A triangle is similar to triangle
. Castruct triangle sach that

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