Question

FIND THE VALUE OF X.
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Answer 1

$\pink\bigstar$Answer:

$\boxed{\sf x \: = {32}^{\circ}}$

$\red\bigstar$Given:

• $\angle$ABC = 46°
• $\angle$EDC = 54°
• $\angle$AED = 132°

$\blue\bigstar$To find:

• The value of x.

$\pink\bigstar$ Solution:

In $\triangle$ACD,

$\because$ $\angle$ACD and $\angle$EDC are opposite interior angles with exterior angle as $\angle$AED,

$\therefore$ $\angle$ACD + $\angle$CDE = $\angle$AED

( Exterior Angle Property Of A Triangle )

$\implies$ $\angle$ACD + 54° = 132°

( Substituting their values )

$\implies$ $\angle$ACD = 132° - 54°

$\implies$ $\boxed{\sf \angle ACD = {78}^{\circ}}$

$\because$ $\angle$ACD + $\angle$ACB = 180°

( Linear Pair )

$\implies$ 78° + $\angle$ACB = 180°

$\implies$ $\angle$ACB = 180° - 78°

$\implies$$\boxed{\sf \angle ACB\: = {102}^{\circ}}$

Now in $\triangle$ABC,

$\because$ $\angle$ABC + $\angle$ACB + $\angle$BAC = 180°

(Angle Sum Property Of A Triangle)

$\implies$ 46° + 102° + x = 180°

( Substituting their values )

$\implies$148° + x = 180°

$\implies$ x = 180° - 148°

$\implies$ $\boxed{\sf x \: = {32}^{\circ}}$

$\sf \boxed{\sf\therefore \: x \: = \: {32}^{ \circ}}$

$\red\bigstar$Concepts Used:

• Exterior Angle Property of a Triangle
• Substitution of values
• Linear pair
• Angle Sum Property Of A Triangle

$\blue\bigstar$Extra - Information:

• Sum of the complementary angles is 90°
• Sum of the supplementary angles is 180°
• Supplementary angles may form a linear pair
• Linear pair of angles are formed when two lines intersect each other at a single point.
• The sum of angles of a linear pair is always equal to 180°.
• Sum of all the interior angles of a Quadrilateral is 360°

Answer 2

Answer:

In \triangle△ ACD,

\because∵ \angle∠ ACD and \angle∠ EDC are opposite interior angles with exterior angle as \angle∠ AED,

\therefore∴ \angle∠ ACD + \angle∠ CDE = \angle∠ AED

( Exterior Angle Property Of A Triangle )

\implies⟹ \angle∠ ACD + 54° = 132°

( Substituting their values )

\implies⟹ \angle∠ ACD = 132° - 54°

\implies⟹ \boxed{\sf \angle ACD = {78}^{\circ}}

∠ACD=78

∘

\because∵ \angle∠ ACD + \angle∠ ACB = 180°

( Linear Pair )

\implies⟹ 78° + \angle∠ ACB = 180°

\implies⟹ \angle∠ ACB = 180° - 78°

\implies⟹ \boxed{\sf \angle ACB\: = {102}^{\circ}}

∠ACB=102

∘

Now in \triangle△ ABC,

\because∵ \angle∠ ABC + \angle∠ ACB + \angle∠ BAC = 180°

(Angle Sum Property Of A Triangle)

\implies⟹ 46° + 102° + x = 180°

( Substituting their values )

\implies⟹ 148° + x = 180°

\implies⟹ x = 180° - 148°

\implies⟹ \boxed{\sf x \: = {32}^{\circ}}

x=32

∘

\sf \boxed{\sf\therefore \: x \: = \: {32}^{ \circ}}

∴x=32

∘

\red\bigstar★ Concepts Used:

Exterior Angle Property of a Triangle

Substitution of values

Linear pair

Angle Sum Property Of A Triangle

\blue\bigstar★ Extra - Information:

Sum of the complementary angles is 90°

Sum of the supplementary angles is 180°

Supplementary angles may form a linear pair

Linear pair of angles are formed when two lines intersect each other at a single point.

The sum of angles of a linear pair is always equal to 180°.

Sum of all the interior angles of a Quadrilateral is 360°