Question

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°.

Answer: ???

Answer 1

The variable, y is 11°

Explanation:

The given parameters are;

in triangle ΔABC;
`{}` in triangle ΔFGH;

Segment
`\overline {AB}` = 14
`{}` Segment
`\overline {FG}` = 14

Segment
`\overline {BC}` = 27
`{}` Segment
`\overline {GH}` = 19

Segment
`\overline {AC}` = 19
`{}` Segment
`\overline {FH}` = 2·y + 5

∡A = 32°
`{}` ∡G = 32°

∡A = ∠BAC which is the angle formed by segments
`\overline {AB}` = 14 and
`\overline {AC}` = 19

Therefore, segment
`\overline {BC}` = 27, is the segment opposite to ∡A = 32°

Similarly, ∡G = ∠FGH which is the angle formed by segments
`\overline {FG}` = 14 and
`\overline {GH}` = 19

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

`\overline {FH}` ≅
`\overline {BC}` by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∴
`\overline {FH}` =
`\overline {BC}` = 27° y definition of congruency

`\overline {FH}` = 2·y + 5 = 27° by transitive property

∴ 2·y + 5 = 27°

2·y = 27° - 5° = 22°

y = 22°/2 = 11°

The variable, y = 11°