In the given figure find the value of $x^{0}$

$60^{\circ}$

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$80^{\circ}$

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$70^{\circ}$

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$90^{\circ}$

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Solution

The correct option is D
$90^{\circ}$

From $Δ$ ADE, sum of all angles of a triangle is $180^{\circ}$ .
So, $∠$ CEB = $70^{\circ}$
From $Δ$ CEB, $∠$ ECB = $90^{\circ}$
$∠$ ECB + $∠$ DCB = $180^{\circ}$
$90^{\circ}$ + $∠$ DCB = $180^{\circ}$
$∠$ DCB = $90^{\circ}$
x = $90^{\circ}$

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Similar questions

Q. Refer to the triangle given in the figure.

x^{0}

in degrees

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∠ A O B = 56^{0}, ∠ B O C = x^{0}

and

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Hint : Sum of all the angles around a point is

360^{0}

In the given figure, it is given that chords $A B = P Q, ∠ A O B = 55^{0},$ and $∠ P O Q = x^{0}$ , find $x$ .

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In the given figure, it is given that chords AB = PQ. The $∠ A O B = 55^{0}$ , and $∠ P O Q = x^{0}$ , find $x$ (in degrees).

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In the given figure find the value of x0

The correct option is D 90∘ From Δ ADE, sum of all angles of a triangle is 180∘. So, ∠CEB = 70∘ From ΔCEB, ∠ECB = 90∘ ∠ECB + ∠DCB = 180∘ 90∘ + ∠DCB = 180∘ ∠DCB = 90∘ x = 90∘

In the given figure find the value of $x^{0}$

The correct option is D
$90^{\circ}$

From $Δ$ ADE, sum of all angles of a triangle is $180^{\circ}$ .
So, $∠$ CEB = $70^{\circ}$
From $Δ$ CEB, $∠$ ECB = $90^{\circ}$
$∠$ ECB + $∠$ DCB = $180^{\circ}$
$90^{\circ}$ + $∠$ DCB = $180^{\circ}$
$∠$ DCB = $90^{\circ}$
x = $90^{\circ}$