In the given figure- m - EDC -54- m - DCA -40- Find x - y and z respectively-A- 20- 27- 86-B- 40- 54- 86-C- 40- 54- 43-D- 20- 27- 43-

The correct option is B 40^∘, 54^∘, 86^∘ m∠ ACD= m ∠ DEC ∴ m∠ DEC=x=40^∘ Similarly, m∠ ECB=m∠ EDC ∴ m∠ ECB=y=54^∘ 54^∘+x+z=180^∘… (sum of all the angles of a t ...

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In the given figure, $m ∠ E D C = 54^{\circ}, m ∠ D C A = 40^{\circ}$ . Find x, y and z respectively.

$20^{\circ}, 27^{\circ}, 86^{\circ}$

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$40^{\circ}, 54^{\circ}, 86^{\circ}$

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$20^{\circ}, 27^{\circ}, 43^{\circ}$

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$40^{\circ}, 54^{\circ}, 43^{\circ}$

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Solution

The correct option is B
$40^{\circ}, 54^{\circ}, 86^{\circ}$

$m ∠$ ACD= m $∠$ DEC
$∴$ m $∠$ DEC=x= $40^{\circ}$
Similarly, m $∠$ ECB=m $∠$ EDC
$∴$ m $∠$ ECB=y= $54^{\circ}$
$54^{\circ} + x + z = 180^{\circ} \dots$ (sum of all the angles of a triangle)
$54^{\circ} + 40^{\circ} + z = 180^{\circ}$
$∴ z = 86^{\circ}$

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In the given figure, m∠EDC=54∘,m∠DCA=40∘. Find x, y and z respectively.

The correct option is B 40∘,54∘,86∘ m∠ ACD= m ∠ DEC ∴ m∠ DEC=x=40∘ Similarly, m∠ ECB=m∠ EDC ∴ m∠ ECB=y=54∘ 54∘+x+z=180∘… (sum of all the angles of a triangle) 54∘+40∘+z=180∘ ∴ z=86∘

In the given figure, $m ∠ E D C = 54^{\circ}, m ∠ D C A = 40^{\circ}$ . Find x, y and z respectively.