In the given figure- l - m - the value of x is205-0285-075-0None

Question

Toppr Admin · Accepted Answer

-x3B8-1-x3B8-2-xk-l-m-x27F9-x3B8-1-40-180-x2218-sumofco-x2212-interiorangles-180-x2218-x27F9-x3B8-1-140-x2218-Similarly-x3B8-2-35-x2218-180-x2218-x27F9-x3B8-2-145-x2218-x2234-x-140-x2218-145-x2218-285-x2218-

In the given figure, if $l | | m$ , the value of x is

$205^{0}$
$285^{0}$
$75^{0}$
None

$θ 1 + θ 2 = x$
$k | | l | | m$
$⟹ θ 1 + 40 = 180^{\circ} (s u m o f c o - i n t e r i o r a n g l e s = 180^{\circ})$
$⟹ θ 1 = 140^{\circ}$
$S i m i l a r l y,$
$θ 2 + 35^{\circ} = 180^{\circ}$
$⟹ θ 2 = 145^{\circ}$
$∴ x = 140^{\circ} + 145^{\circ} = 285^{\circ}$

In the given figure, if l||m, the value of x is20502850750None

θ1+θ2=xk||l||m⟹θ1+40=180∘(sumofco−interiorangles=180∘)⟹θ1=140∘Similarly,θ2+35∘=180∘⟹θ2=145∘∴x=140∘+145∘=285∘

In the given figure, if $l | | m$ , the value of x is

$205^{0}$
$285^{0}$
$75^{0}$
None

$θ 1 + θ 2 = x$
$k | | l | | m$
$⟹ θ 1 + 40 = 180^{\circ} (s u m o f c o - i n t e r i o r a n g l e s = 180^{\circ})$
$⟹ θ 1 = 140^{\circ}$
$S i m i l a r l y,$
$θ 2 + 35^{\circ} = 180^{\circ}$
$⟹ θ 2 = 145^{\circ}$
$∴ x = 140^{\circ} + 145^{\circ} = 285^{\circ}$