In the given figure- AB - DC and AD - BC-Also- - ADC - 2x-5- and - BCD - 3x-55- Find the measure of x-

AB ∥ DC and AD ∥ BC ∠ ADC + ∠ BCD = 180^∘ ( Sum of Co interior angles) = gt; (2x+5) + nbsp;(3x+55) = 180 = (2x+3x) + nbsp;(5+55) = 180 = (5x) + nb ...

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Byju's Answer

Standard VII

Mathematics

Special Names for Pairs of Angles Made by Parallel Lines and Transversal

In the given ...

Question

In the given figure, $A B ∥ D C a n d A D ∥ B C$ .

Also, $∠ A D C = (2 x + 5)^{\circ} a n d ∠ B C D = (3 x + 55)^{\circ}$ . Find the measure of x.

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Solution

$A B ∥ D C a n d A D ∥ B C$

$∠ A D C + ∠ B C D = 180^{\circ} (S u m o f C o - i n t e r i o r a n g l e s)$

=> $(2 x + 5) + (3 x + 55) = 180$

= $(2 x + 3 x) + (5 + 55) = 180$

= $(5 x) + (60) = 180$

= $5 x = 180 - 60$

= $5 x = 120$

=> $x = 24^{\circ}$

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In the given figure, AB∥DC and AD∥BC. Also, ∠ADC=(2x+5)∘ and ∠BCD=(3x+55)∘. Find the measure of x.

AB∥DC and AD∥BC ∠ADC+∠BCD=180∘ (Sum of Co−interior angles) => (2x+5)+(3x+55)=180 = (2x+3x)+(5+55)=180 = (5x)+(60)=180 = 5x=180−60 = 5x=120 => x=24∘

In the given figure, $A B ∥ D C a n d A D ∥ B C$ .

Also, $∠ A D C = (2 x + 5)^{\circ} a n d ∠ B C D = (3 x + 55)^{\circ}$ . Find the measure of x.

$A B ∥ D C a n d A D ∥ B C$

$∠ A D C + ∠ B C D = 180^{\circ} (S u m o f C o - i n t e r i o r a n g l e s)$

=> $(2 x + 5) + (3 x + 55) = 180$

= $(2 x + 3 x) + (5 + 55) = 180$

= $(5 x) + (60) = 180$

= $5 x = 180 - 60$

= $5 x = 120$

=> $x = 24^{\circ}$