The correct option is A 50^∘We know that angle in a semicircle is 90^∘. ∴∠ ABC=90^∘ Using angle sum property in ABC, we have ∠ ABC+∠ ACB+∠ BAC=180^∘. 90^∘+40^∘ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Angle Subtended by an Arc of a Circle at the Centre

Question

If $∠ A C B = 40^{\circ},$ then what is the value of $x$ ?

50^{\circ}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

90^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $50^{\circ}$
We know that angle in a semicircle is $90^{\circ} .$
$∴ ∠ A B C = 90^{\circ}$
Using angle sum property in $△ A B C,$ we have
$∠ A B C + ∠ A C B + ∠ B A C = 180^{\circ} .$
$⟹ 90^{\circ} + 40^{\circ} + x = 180^{\circ}$
$x = 180^{\circ} - 130^{\circ} = 50^{\circ}$

Suggest Corrections

Similar questions

Q. Prove that

(i) s e c^{2} θ + c o s e c^{2} θ = s e c^{2} θ c o s e c^{2} θ

(i i) t a n^{2} θ - s i n^{2} θ = t a n^{2} θ s i n^{2} θ

(i i i) t a n^{2} θ + c o t^{2} θ + 2 = s e c^{2} θ c o s e c^{2} θ

[3 MARKS]

Is LHS=RHS?

\sqrt{\frac{csc θ - cot θ}{csc θ + cot θ}} + \sqrt{\frac{csc θ + cot θ}{csc θ - cot θ}} = 2 csc θ

Say true or false.

sin (- θ) + cos (- θ) + sec (- θ) + sin (π - θ) + cos (π - θ) + sec (π - θ) =

Q. If

csc θ - cot θ = \sqrt 2 . cot θ

, then prove that

csc θ + cot θ = \sqrt 2 . csc θ

Q. If

f (θ) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {cos}^{2} θ & cos θ sin θ & - sin θ cos θ sin θ & {sin}^{2} θ & cos θ sin θ & - cos θ & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

then for all

θ

Join BYJU'S Learning Program

If ∠ACB=40∘, then what is the value of x?

The correct option is A 50∘We know that angle in a semicircle is 90∘. ∴∠ABC=90∘ Using angle sum property in △ABC, we have ∠ABC+∠ACB+∠BAC=180∘. ⟹90∘+40∘+x=180∘ x=180∘−130∘=50∘

If $∠ A C B = 40^{\circ},$ then what is the value of $x$ ?