If we have -ABC as - reflex -ABC is equal to - nbsp-180- - nbsp-180- - nbsp-360- - nbsp-360- - nbsp-

The correct option is C 360^∘ θ nbsp; From the figure, we can see that ∠ ABC + Reflex nbsp;∠ ABC = 360^∘ So, Reflex ∠ ABC = 360^∘ θ ...

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Byju's Answer

Standard VI

Mathematics

Acute, Obtuse and Reflex Angles

If we have ∠A...

Question

If we have ∠ABC as θ, reflex ∠ABC is equal to _______.

$180^{\circ} - θ$

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$- 180^{\circ} - θ$

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$360^{\circ} - θ$

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$360^{\circ} - θ$

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Solution

The correct option is C
$360^{\circ} - θ$

From the figure, we can see that $∠ A B C +$ Reflex $∠ A B C = 360^{\circ}$

So, Reflex $∠ A B C = 360^{\circ} - θ$

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Similar questions

Q. Prove that
(i)

\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1

(ii)

\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2

(iii)

\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1

(iv)

\{1 + cotθ - \sec (\frac{π}{2} + θ)\} \{1 + cotθ + \sec (\frac{π}{2} + θ)\} = 2 cotθ

(v)

\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1

Prove that:

$(i) \frac{c o s (2 π + θ) c o s e c (2 π + θ) t a n (π / 2 + θ)}{s e c (π / 2 + θ) c o s θ c o t (π + θ)} = 1$

$(i i) \frac{c o s e c (90^{\circ} + θ) + c o t (450^{\circ} + θ)}{c o s e c (90^{\circ} - θ) + t a n (180^{\circ} - θ)} + \frac{t a n (180^{\circ} + θ) + s e c (180^{\circ} - θ)}{t a n (360^{\circ} + θ) - s e c (- θ)} = 2$

$(i i i) \frac{s i n (180^{\circ} + θ) c o s (90^{\circ} + θ) t a n (270^{\circ} - θ) c o t (360^{\circ} - θ)}{s i n (360^{\circ} - θ) c o s (360^{\circ} + θ) c o s e c (- θ) s i n (270^{\circ} + θ)} (i v) 1 + c o t θ - s e c (\frac{π}{2} + θ)} 1 + c o t θ + s e c (\frac{π}{2} + θ)} = 2 c o t θ$

$(v) \frac{t a n (90^{\circ} - θ) s e c (180^{\circ} - θ) s i n (- θ)}{s i n (180^{\circ} + θ) c o t (360^{\circ} - θ) c o s e c (90^{\circ} - θ)} = 1$

If we have $∠ A B C$ as $θ$ , reflex $∠ A B C$ is equal to ________________.

If we have $∠$ ABC as $t h e t a$ , reflex $a n g l e$ ABC is equal to ________________.

If we have ∠ABC as θ, reflex ∠ABC is equal to _______.

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If we have ∠ABC as θ, reflex ∠ABC is equal to _______.

The correct option is C 360∘−θ From the figure, we can see that ∠ABC+ Reflex ∠ABC=360∘ So, Reflex ∠ABC=360∘−θ ​​

The correct option is C
$360^{\circ} - θ$

From the figure, we can see that $∠ A B C +$ Reflex $∠ A B C = 360^{\circ}$

So, Reflex $∠ A B C = 360^{\circ} - θ$