In the figure below- ABC - BED - - DEB -115- and - CAB -25- find - BDE-A- 40-B- 45-C- 50-D- 55-

The correct option is A 40° Since ABC ≅ BED ∠DEB = ∠CBA = 115^∘ ∠CAB = ∠DBE = 25^∘ ∠CBA,∠DBE and ∠CBD from a linear pair. ∠CBA + ∠DBE + ∠CBD = 180^∘ So, ∠CBD ...

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Byju's Answer

Standard IX

Mathematics

Triangle and Sum of Its Internal Angles

In the figure...

Question

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ , find $∠$ BDE.

40°

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45°

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50°

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55°

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Solution

The correct option is A
40°

Since $△$ ABC $≅$ $△$ BED

$∠$ DEB = $∠$ CBA = $115^{\circ}$
$∠$ CAB = $∠$ DBE = $25^{\circ}$
$∠$ CBA, $∠$ DBE and $∠$ CBD from a linear pair.

$∠$ CBA + $∠$ DBE + $∠$ CBD = $180^{\circ}$
So, $∠$ CBD = $40^{\circ}$

$∠$ CBE + $∠$ BED = $180^{\circ}$
So, BC $∥$ ED

$∠$ BDE = $∠$ CBD = $40^{\circ}$ (Alternate interior angles)

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Similar questions

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ . It can be concluded that BC || ED.

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ . It can be concluded that BC||ED.

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ . It can be concluded that

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ . It can be concluded that BC||ED.

Q. In the given figure, ∠BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?
(a) 50°
(b) 30°
(c) 40°
(d) 25°

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In the figure below, △ABC ≅ △BED, ∠DEB = 115∘ and ∠CAB = 25∘, find ∠BDE.

The correct option is A 40° Since △ABC ≅ △BED ∠DEB = ∠CBA = 115∘ ∠CAB = ∠DBE = 25∘ ∠CBA,∠DBE and ∠CBD from a linear pair. ∠CBA + ∠DBE + ∠CBD = 180∘ So, ∠CBD = 40∘ ∠CBE + ∠BED =180∘ So, BC ∥ ED ∠BDE = ∠CBD = 40∘ (Alternate interior angles)

In the figure below, $△$ ABC $≅$ $△$ BED, $∠$ DEB = $115^{\circ}$ and $∠$ CAB = $25^{\circ}$ , find $∠$ BDE.