Find the value of the variable. In triangle ABC, AB¯¯¯¯¯¯¯¯=14, BC¯¯¯¯¯¯¯¯=27, AC¯¯¯¯¯¯¯¯=19, and ∡A=32°. In triangle FGH, FG¯¯¯¯¯¯¯¯=14, GH¯¯¯¯¯¯¯¯=19, FH¯¯¯¯¯¯¯¯=2y+5, and ∡G=32°.

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The variable, y is 11°

Step-by-step explanation:

The given parameters are;

in triangle ΔABC;                        in triangle ΔFGH;

Segment = 14                       Segment = 14

Segment = 27                      Segment = 19

Segment = 19                       Segment = 2·y + 5

∡A = 32°                                      ∡G = 32°

∡A = ∠BAC which is the angle formed by segments = 14 and = 19

Therefore, segment = 27, is the segment opposite to ∡A = 32°

Similarly, ∡G = ∠FGH which is the angle formed by segments = 14 and = 19

Therefore, segment = 2·y + 5, is the segment opposite to ∡A = 32° and triangle ΔABC ≅ ΔFGH by Side-Angle-Side congruency rule which gives;

≅ by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∴ = = 27° y definition of congruency

= 2·y + 5 = 27° by transitive property

∴ 2·y + 5 = 27°

2·y = 27° - 5° = 22°

y = 22°/2 = 11°

The variable, y = 11°

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )