In the given figure- z -A- 110-B- 45-C- None of theseD- 40-

The correct option is B 45° ∠ABC = 180^∘ – 110^∘ = 70^∘ ∠ADB = ∠ABD =70^∘ ...... (given, AB = AD) Also, ∠ADB = ∠DCA + ∠DAC ....... (exterior angle theorem) ...

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Standard VI

Mathematics

Classification of Triangles Based on Sides

In the given ...

Question

In the given figure, z =___.

40°

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110°

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45°

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None of these

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Solution

The correct option is C
45°

$∠$ ABC = $180^{\circ}$ – $110^{\circ}$ = $70^{\circ}$

$∠$ ADB = $∠$ ABD = $70^{\circ}$ ...... (given, AB = AD)

Also, $∠$ ADB = $∠$ DCA + $∠$ DAC ....... (exterior angle theorem)

70° = 25° + z°

z° = 45°

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In the given figure, z =___.

The correct option is C 45° ∠ABC = 180∘ – 110∘ = 70∘ ∠ADB = ∠ABD =70∘ ...... (given, AB = AD) Also, ∠ADB = ∠DCA + ∠DAC ....... (exterior angle theorem) 70° = 25° + z° z° = 45°

The correct option is C
45°

$∠$ ABC = $180^{\circ}$ – $110^{\circ}$ = $70^{\circ}$

$∠$ ADB = $∠$ ABD = $70^{\circ}$ ...... (given, AB = AD)

Also, $∠$ ADB = $∠$ DCA + $∠$ DAC ....... (exterior angle theorem)

70° = 25° + z°

z° = 45°