Open the file and please answer the question below.
Find the measure of the lettered angles:
∠DAE = (x+19)°, ∠BAC = (2x+y)°, ∠ABC = (y-11)°, ∠ACB = (x+11)°
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Answered by
1
I think you forgot to mention that AE and BC are parallel. I am assuming they are parallel.
If AE and BC are parallel,
∠DAE = ∠ABC
⇒ x+19 = y - 11
⇒ x = y - 11 - 19
⇒ x = y - 30
∠BAC, ∠ABC and ∠ACB are the three angles of a triangle. Sum of these angles is 180°.
(2x+y)°+ (y-11)°+ (x+11)° = 180°
⇒ 2x+y + y - 11 + x + 11 = 180
⇒ 3x + 2y = 180
⇒ 3(y-30) + 2y = 180
⇒ 5y - 90 = 180
⇒ 5y = 270
⇒ y = 270/5 = 54°
x = y - 30 = 24°
∠DAE = 24+19 = 43°
∠BAC = 2×24+54=102°
∠ABC = 54-11 = 43°
∠ACB = 24+11 = 35°
If AE and BC are parallel,
∠DAE = ∠ABC
⇒ x+19 = y - 11
⇒ x = y - 11 - 19
⇒ x = y - 30
∠BAC, ∠ABC and ∠ACB are the three angles of a triangle. Sum of these angles is 180°.
(2x+y)°+ (y-11)°+ (x+11)° = 180°
⇒ 2x+y + y - 11 + x + 11 = 180
⇒ 3x + 2y = 180
⇒ 3(y-30) + 2y = 180
⇒ 5y - 90 = 180
⇒ 5y = 270
⇒ y = 270/5 = 54°
x = y - 30 = 24°
∠DAE = 24+19 = 43°
∠BAC = 2×24+54=102°
∠ABC = 54-11 = 43°
∠ACB = 24+11 = 35°
Answered by
1
the sum of angles in triangle ABC = 180 deg =>
angle BAC + angle ABC + angle ACB = 180°
2x+y + y - 11 + x + 11 = 180°
3x + 2y = 180° --- (1)
Angle DAE = angle ABC as AE parallel to BC. DAB is the transversal line.
x+19 = y - 11
y - x = 30° --- (2)
(1) + 3 * (2) => 5 y = 270° => y = 54°
using (2) we get : x = 24°
Now the angles are
∠DAE = x+19° = 43°
∠BAC = 2x+y = 102°
∠ ABC = y - 11° = 43°
∠ACB = x + 11° = 35°
angle BAC + angle ABC + angle ACB = 180°
2x+y + y - 11 + x + 11 = 180°
3x + 2y = 180° --- (1)
Angle DAE = angle ABC as AE parallel to BC. DAB is the transversal line.
x+19 = y - 11
y - x = 30° --- (2)
(1) + 3 * (2) => 5 y = 270° => y = 54°
using (2) we get : x = 24°
Now the angles are
∠DAE = x+19° = 43°
∠BAC = 2x+y = 102°
∠ ABC = y - 11° = 43°
∠ACB = x + 11° = 35°
kvnmurty:
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