In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then $∠ C A D + ∠ C B D =$ ?

120^{\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!

360^{\circ}

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

180^{\circ}

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

Open in App

Solution

The correct option is D $180^{\circ}$
$∠ C A B = ∠ B C D a n d ∠ D A B = ∠ B D C ∴ ∠ C A D = ∠ C A B + ∠ D A B = ∠ B C D + ∠ B D C \Rightarrow ∠ C A D = ∠ B C D + ∠ B D C ∴ ∠ C A D + ∠ C B D = ∠ B C D + ∠ B D C + ∠ C B D = 180^{\circ}$

Suggest Corrections

Similar questions

Q. In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then

∠ C A D + ∠ C B D = . . . . . . . . . .

Q. Two circles intersect each other at A and B.Let DC be a common tangent touching the circle in point C and D
Prove that

∠ CAD + ∠ CBD = 180 °

In the figure, AB is the common tangent to the 2 circles intersecting at C & D. Write down the numerical value of ∠ACB + ∠ADB. Give reasons for your answer.

Q. In the given figure, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove that ▢ABCD is cyclic.

Two circles touch each other externally at point P in the given figure. AB is the direct common tangent of these circles. Then ÐAPB equals to:

Join BYJU'S Learning Program

In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then ∠CAD+∠CBD=?

The correct option is D 180∘∠CAB=∠BCDand ∠DAB=∠BDC∴ ∠CAD=∠CAB+∠DAB=∠BCD+∠BDC⇒ ∠CAD=∠BCD+∠BDC∴ ∠CAD+∠CBD=∠BCD+∠BDC+∠CBD=180∘

In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then $∠ C A D + ∠ C B D =$ ?

The correct option is D $180^{\circ}$
$∠ C A B = ∠ B C D a n d ∠ D A B = ∠ B D C ∴ ∠ C A D = ∠ C A B + ∠ D A B = ∠ B C D + ∠ B D C \Rightarrow ∠ C A D = ∠ B C D + ∠ B D C ∴ ∠ C A D + ∠ C B D = ∠ B C D + ∠ B D C + ∠ C B D = 180^{\circ}$