Construct a triangle ABC with AB 5 cm and angle A measures 45° and angle C measures 120°

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Please find below the solution to the asked query:

Given to construct triangle ABC ,  AB  =  5 cm
And
$\angle$ BAC  =  45$°$  , $\angle$ ACB  =  120$°$

And we know from angle sum property of triangle

$\angle$ BAC  + $\angle$ ACB  + $\angle$ ABC  =  180$°$   , So

45$°$ +  120$°$ + $\angle$ ABC  =  180$°$ 

$\angle$ ABC  =  15$°$ 

Now we follow these steps to construct triangle ABC , As :

Step 1 :  Draw a line AB  =  5 cm

Step 2 :  Take any radius (  Less than half of AB ) and center " A "  we draw a semicircle that intersect our line AB at " P " . With same radius take center "  P "  draw an arc that intersect our semicircle at " Q " And with same radius take center "  Q "  draw an arc that intersect our semicircle at " R " .

Step 3 :  With same radius take center "  Q "  and " R " draw two arcs that intersect each other at  " S " .

Step 4 :  Join line AS , that intersect our semicircle at " T "

Step 5 :  With same radius take center "  T "  and " P " draw two arcs that intersect each other at  " U "  . Join AU  .

Step 6 : Take any radius (  Less than half of AB ) and center " B "  we draw a semicircle that intersect our line AB at " V " . With same radius take center "  V "  draw an arc that intersect our semicircle at " W " .

Step 7 :  With same radius take center "  V  "  and " W " draw two arcs that intersect each other at  " X " .

Step 8 :  Join BX and BX intersect our semicircle at " Y " .

Step 9 :  With same radius take center "  Y "  and " V " draw two arcs that intersect each other at  " Z " join BZ and extend it , and that intersect line AU at " C "  , We get triangle ABC .



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