Question

In the given figure, p and q are two
parallel lines. AB and BC are line segments
such that ZBAD = 140° and _BCE = 120°
Find the measures of the reflex ABC and
the smaller ZABC at B. You may have to
do a small construction to proceed with
your proof. Think what you have to do​

Answer 1

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✤ Required Answer:

✒ GiveN:

• ∠BAD = 140°
• ∠BCE = 120°

✒ To FinD:

• Find ∠ABC.....?
• And, reflex ∠ABC...?

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✤ How to solve?

For solving this question, we need to know how the angles are classified when a transverse cuts two parallel lines.

• Internal angles on same of transversal- When a transdermal cuts two parallel, the angle formed on the same side internally adds upto 180°
• Around a point, The angles add upto 360°.

☃️ So, By using the above concept, Let's solve this question....$.$

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✤ Solution:

Construction:

• Draw a line parallel with line p and q from point B. And let a point on that line X

Then,

• AB is like a transversal for line p and BX and BC is like a transversal for line q and BX

Here,

• ∠BAD and ∠ABX adds upto 180°

➝ ∠BAD + ∠ABX = 180°

➝ 140° + ∠ABX = 180°

➝ ∠ABX = 180° - 140°

➝ ∠ABX = 40°

And,

• ∠CBX and ∠BCE adds upto 180°

➝ ∠CBX + ∠BCE = 180°

➝ ∠CBX + 120° = 180°

➝ ∠CBX = 180° - 120°

➝ ∠CBX = 60°

We had to find ∠ABC

➝ ∠ABC = ∠ABX + ∠CBX

➝ ∠ABC = 40° + 60°

➝ ∠ABC = 100°

✒ Therefore, ∠ABC = 100°

Around point B,

➝ ∠ABC + reflex ∠ABC = 360°

➝ 100° + reflex ∠ABC = 360°

➝ reflex ∠ABC = 360° - 100°

➝ reflex ∠ABC = 260°

✒ And also, reflex ∠ABC = 260°

☀️ Hence, solved !!

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Answer 2

$\large\:\sf\underline{\red{SOLUTION:-}}$

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♻️ See the attachment figure .

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Required drawing :-

• Draw a line ‘r’, i.e parallel to lines ‘p’ and ‘q’ and take two points ‘x’ and ‘y’ as shown in figure .

Properties of parallel lines :-

• Alternative interior angles between two parallel lines are equal to each other .

(✧) So, here

$\rm{\green{\angle{DAB}\:=\:\angle{ABx}\:=\:140^{\degree}\:}}$

$\rm{\red{\angle{ECB}\:=\:\angle{CBx}\:=\:120^{\degree}\:}}$

$\rm{\purple{\angle{pAB}\:=\:\angle{ABy}\:=\:40^{\degree}\:}}$

$\rm{\blue{\angle{qCB}\:=\:\angle{CBy}\:=\:60^{\degree}\:}}$

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CALCULATION :-

✪ Measurement of Reflex ABC is,

$\rm{=\:\angle{ABx}\:+\:\angle{CBx}\:}$

$\rm{=\:140^{\degree}\:+\:120^{\degree}\:}$

$\rm{=\:260^{\degree}\:}$

✪ Measurement of smaller $\rm{\angle{ABC}\:}$ is,

$\rm{=\:\angle{ABy}\:+\:\angle{CBy}\:}$

$\rm{=\:40^{\degree}\:+\:60^{\degree}\:}$

$\rm{=\:100^{\degree}\:}$