39. A quadrilateral EFGH is divided by its diagonal EG into two triangles, EFG and EHG. If the angles of triangle EFG are 50°, 60°, and 70°, and one angle of triangle EHG is 40°, what is the measure of the smallest angle in quadrilateral EFGH?
Key Concept: Sum of angles in a quadrilateral, diagonals creating triangles
b) 40°
[Solution Description]
Step 1: In triangle EFG, angles are 50°, 60°, and 70° (sum = 180°).
Step 2: Diagonal EG divides the quadrilateral into two triangles, EFG and EHG.
Step 3: Given one angle of triangle EHG is 40°. Let the other two angles be x and y ⇒ $40° + x + y = 180°$ ⇒ $x + y = 140°$.
Step 4: The quadrilateral's angles are the combination of both triangles' angles: 50°, 60°, 70°, 40°, x, y.
But since the quadrilateral has only four angles, they must be 50°, 60°, 70°, and 40° (the unique angles from both triangles). However, this leads to a sum of $50° + 60° + 70° + 40° = 220° ≠360°$, which is incorrect.
Upon reevaluating, the quadrilateral's angles are formed by combining adjacent angles from the two triangles. For example, $∠FEG + ∠HEG$ forms angle E, and similarly for others. Therefore, the quadrilateral's angles are:
$∠E = ∠FEG + ∠HEG$, $∠F = ∠EFG$, $∠G = ∠EGF + ∠EGH$, $∠H = ∠EHG$.
Given $∠F = 60°$ (from triangle EFG), $∠H = 40°$ (given in EHG), $∠FEG = 50°$, $∠EGF = 70°$, and let $∠HEG = z$, $∠EGH = w$.
Since $∠FEG + ∠HEG = ∠E ⇒ 50° + z = ∠E$.
And $∠EGF + ∠EGH = ∠G ⇒ 70° + w = ∠G$.
Also, in triangle EHG: $40° + z + w = 180° ⇒ z + w = 140°$.
Now, sum of quadrilateral angles: $∠E + ∠F + ∠G + ∠H = 360° ⇒ (50° + z) + 60° + (70° + w) + 40° = 360° ⇒ 220° + z + w = 360° ⇒ z + w = 140°$.
This confirms consistency but doesn't provide new information. To find the smallest angle, note that the quadrilateral's angles include $∠F = 60°$, $∠H = 40°$, and $∠E$ and $∠G$ are combinations of angles greater than or equal to their components. Thus, the smallest angle is $∠H = 40°$.
Your Answer is correct.
b) 40°
[Solution Description]
Step 1: In triangle EFG, angles are 50°, 60°, and 70° (sum = 180°).
Step 2: Diagonal EG divides the quadrilateral into two triangles, EFG and EHG.
Step 3: Given one angle of triangle EHG is 40°. Let the other two angles be x and y ⇒ $40° + x + y = 180°$ ⇒ $x + y = 140°$.
Step 4: The quadrilateral's angles are the combination of both triangles' angles: 50°, 60°, 70°, 40°, x, y.
But since the quadrilateral has only four angles, they must be 50°, 60°, 70°, and 40° (the unique angles from both triangles). However, this leads to a sum of $50° + 60° + 70° + 40° = 220° ≠360°$, which is incorrect.
Upon reevaluating, the quadrilateral's angles are formed by combining adjacent angles from the two triangles. For example, $∠FEG + ∠HEG$ forms angle E, and similarly for others. Therefore, the quadrilateral's angles are:
$∠E = ∠FEG + ∠HEG$, $∠F = ∠EFG$, $∠G = ∠EGF + ∠EGH$, $∠H = ∠EHG$.
Given $∠F = 60°$ (from triangle EFG), $∠H = 40°$ (given in EHG), $∠FEG = 50°$, $∠EGF = 70°$, and let $∠HEG = z$, $∠EGH = w$.
Since $∠FEG + ∠HEG = ∠E ⇒ 50° + z = ∠E$.
And $∠EGF + ∠EGH = ∠G ⇒ 70° + w = ∠G$.
Also, in triangle EHG: $40° + z + w = 180° ⇒ z + w = 140°$.
Now, sum of quadrilateral angles: $∠E + ∠F + ∠G + ∠H = 360° ⇒ (50° + z) + 60° + (70° + w) + 40° = 360° ⇒ 220° + z + w = 360° ⇒ z + w = 140°$.
This confirms consistency but doesn't provide new information. To find the smallest angle, note that the quadrilateral's angles include $∠F = 60°$, $∠H = 40°$, and $∠E$ and $∠G$ are combinations of angles greater than or equal to their components. Thus, the smallest angle is $∠H = 40°$.