Find - BOA-A- 150-B- 100-C- 80-D- Indeterminate

The correct option is B 100^∘∠ CAF= 100^∘. Hence ∠ BAC =80^∘ Also, ∠ OCA=(90 ∠ ACF)=90^∘ 50^∘=40^∘=∠ OAC (Since the triangle OCA is isosceles) Hence ∠ OAB=40^ ...

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Quantitative Aptitude

Circles

Find ∠ BOA.A....

Question

Find $∠ B O A$ .

100^{\circ}

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150^{\circ}

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80^{\circ}

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Indeterminate

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Solution

The correct option is A $100^{\circ}$
$∠ C A F = 100^{\circ} . H e n c e ∠ B A C = 80^{\circ}$
Also, $∠ O C A = (90 - ∠ A C F) = 90^{\circ} - 50^{\circ} = 40^{\circ} = ∠ O A C$
(Since the triangle OCA is isosceles)
Hence $∠ O A B = 40^{\circ}$
In isosceles $Δ O A B, ∠ O B A$ will also be $40^{\circ}$
Hence, $∠ B O A = 180^{\circ} - 40^{\circ} - 40^{\circ} = 100^{\circ}$

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Similar questions

In the given figure, AB||CD. If $∠ A O C = 30^{\circ}$ and $∠ O A B = 100^{\circ}$ then $∠ O C D = ?$

$(a) 130^{\circ}$

$(b) 150^{\circ}$

$(c) 80^{\circ}$

$(d) 100^{\circ}$

Q. If ∆PQR ≅ ∆ABC and ∠B = ∠C = 50°, then find the measure of ∠P.

Q. In the figure given below (not drawn to scale), A,B,C are three points on a circle with center O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If

∠ A T C = 30^{\circ}

and

∠ A C T = 50^{\circ}

. Find

∠ B O A

The circumference of a circle with centre O is divided into three arcs APB, BQC and CRA such that

$\frac{a r c A P B}{2} = \frac{a r c B Q C}{3} = \frac{a r c C R A}{4}$ . Find $∠ B O A .$

Q. In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°

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Find ∠BOA.

The correct option is A 100∘∠CAF=100∘.Hence ∠BAC=80∘ Also, ∠OCA=(90−∠ACF)=90∘−50∘=40∘=∠OAC (Since the triangle OCA is isosceles) Hence ∠OAB=40∘ In isosceles ΔOAB,∠OBA will also be 40∘ Hence, ∠BOA=180∘−40∘−40∘=100∘

Find $∠ B O A$ .