find the unknown angle in the following figure
Answers
❉ Given:-
1. In triangle₁ ∆ABC
- AB = AC
- ∠ BAC = 30°
2. In triangle₂ ∆ACD and ∆ABD
- DB = AB = BC
- ∠ BAD = 40°
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❉ To Find:-
In triangle₁ ∆ABC
- Value of y = ∠ DBC
In triangle₂ ∆ACD
- ∆ACD = ∆ADB + ∆BCD
- Value of z = ∠BDA
- Value of x = ∠CBD
- Value of y = ∠BDC
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❉ Property Used:-
- Sum of angles of triangle is 180°
- Isosceles triangle has two equal sides .
- Isosceles triangle equal sides substend equal angles.
- Sum of angles in a straight line is 180° known as linear pair.
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❉ Solution:-
✦ Finding the value of y in triangle₁ ∆ABC
In triangle₁ ∆ABC
AB = BC [ Given ]
➥ As we know that triangle with two equal angles is an isosceles triangle;
➥ According to property of isosceles triangle
∠ABC = ∠ ACB -------- eq.1
➥ As we know,
Sum of angles of triangle = 180°
➟ ∠ABC + ∠ ACB + ∠ BAC = 180°
[ Putting eq. 1 in place of ∠ ACB]
➟ ∠ABC + ∠ ABC + 30 = 180°
➟ 2∠ABC + 30 = 180
➟ 2∠ABC = 180 - 30
➟ ∠ABC = 150 ÷ 2
➟ ∠ABC = 75°
∴∠ABC = 75°
➥ According to given conditions
∠ABC + ∠DBC = 180 [ By linear pair ]
➟ ∠ABC + y = 180
➟ 75 + y = 180
➟ y = 180 - 75
➟ y = 105
∴ Value of y is 105°
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✦ Finding the value of z in triangle₂ ∆ACD
In ∆ABD
AB = BD
➥ As we know that triangle with two equal angles is an isosceles triangle;
➥ According to property of isosceles triangle
∠BAD = ∠BDA
Hence,
z = ∠BDA = ∠BAD = 40°
∴ Value of z is 40°
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✦ Finding the value of x in triangle₂ ∆ACD
➥ As we know,
Sum of angles of triangle = 180°
➟ ∠ABD + ∠ BDA + ∠ BAD = 180°
➟ ∠ABD + 40 + 40 = 180°
➟ ∠ABD + 80 = 180
➟ ∠ABD = 180 - 80
➟ ∠ABD = 100°
∴ ∠ABD = 100°
➥ According to given conditions;
∠ABD + ∠DBC = 180° [ By linear pair ]
➟ ∠ABD + x = 180°
➟ 100 + x = 180°
➟ x = 180 - 100
➟ x = 80
∴ Value of x is 80°
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✦ Finding the value of y in triangle₂ ∆ACD
In triangle₁ ∆DBC
DB = BC [ Given ]
➥ As we know that triangle with two equal angles is an isosceles triangle;
➥ According to property of isosceles triangle
∠BDC = ∠ BCD -------- eq.2
➥ As we know,
Sum of angles of triangle = 180°
➟ ∠BDC + ∠ BCD + ∠ DBC = 180°
[ Putting eq. 2 in place of ∠ BCD]
➟ ∠BDC + ∠ BDC + 80° = 180°
➟ 2∠BDC + 80 = 180
➟ 2∠BDC = 180 - 80
➟ ∠BDC = 100 ÷ 2
➟ ∠BDC = 50°
∠BDC = 50°
∴ Value of y is ∠BDC = 50°
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❉ Answer:-
- Value of y in triangle₁ ∆ABC is 105°
- Value of z in triangle₂ ∆ACD is 40°
- Value of x in triangle₂ ∆ACD is 80°
- Value of y in triangle₂ ∆ACD is 50°
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