In Fig- 73 - the value of x isa 20b 30c 40d 25P

∠TRS + ∠TRQ = 180° nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; [Linear angles] ⇒ 5x° + nbsp;∠TRQ = 180° ⇒ ∠TRQ = 180° − nbsp;5x° Now, ∠Q ...

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Byju's Answer

Standard IX

Mathematics

Exterior Angle of a Triangle = Sum of Opposite Internal Angles

In Fig. 73 , ...

Question

In Fig. 73, the value of x is
(a) 20
(b) 30
(c) 40
(d) 25

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Solution

∠TRS + ∠TRQ = 180° [Linear angles]
⇒ 5x° + ∠TRQ = 180°
⇒ ∠TRQ = 180° − 5x°
Now, ∠QTR + ∠TRQ = ∠PQT [Exterior angle property of triangle]
⇒ 3x° + 180° − 5x° = 120°
⇒ 2x° = 60°
⇒ x = 30
Hence, the correct answer is option (b).

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In Fig. 73, the value of x is (a) 20 (b) 30 (c) 40 (d) 25

∠TRS + ∠TRQ = 180° [Linear angles] ⇒ 5x° + ∠TRQ = 180° ⇒ ∠TRQ = 180° − 5x° Now, ∠QTR + ∠TRQ = ∠PQT [Exterior angle property of triangle] ⇒ 3x° + 180° − 5x° = 120° ⇒ 2x° = 60° ⇒ x = 30 Hence, the correct answer is option (b).

In Fig. 73, the value of x is
(a) 20
(b) 30
(c) 40
(d) 25

∠TRS + ∠TRQ = 180° [Linear angles]
⇒ 5x° + ∠TRQ = 180°
⇒ ∠TRQ = 180° − 5x°
Now, ∠QTR + ∠TRQ = ∠PQT [Exterior angle property of triangle]
⇒ 3x° + 180° − 5x° = 120°
⇒ 2x° = 60°
⇒ x = 30
Hence, the correct answer is option (b).