C-08-Q171: In the proof of Theorem 26, $\triangle PAO$ and $\triangle PBO$ are— | প্রশ্ন করি

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C-08-Q171mediumsingle_mcq

In the proof of Theorem 26, $\triangle PAO$ and $\triangle PBO$ are—

aequilateral
bright-angled at $A$ and $B$ respectively
cisosceles only
dobtuse-angled

ব্যাখ্যা

Each tangent is perpendicular to the radius at its point of contact, so $\angle OAP$ and $\angle OBP$ are each $90^{\circ}$ . Thus the triangles $\triangle PAO$ and are right-angled at and .

171/231

এই অধ্যায়ের আরও প্রশ্ন

C-08-Q172The congruence used in Theorem 26 is—
C-08-Q173After the congruence, because they are—
C-08-Q174If and are tangents from an external point to a circle with centre , then is—
C-08-Q175also bisects the angle—
C-08-Q176Two tangents and are drawn from to a circle. If and centre is , then equals—
C-08-Q177Two tangents and are drawn from to a circle with centre such that . The value of is—
C-08-Q178Theorem 27: if two circles touch each other externally, the point of contact and the two centres are—
C-08-Q179The proof of Theorem 27 uses the fact that at the point of contact—
C-08-Q180Since right angle, at the point of contact is a—

△ P A O

\triangle PBO

A

B

C-08-Q181Corollary 11: if two circles touch each other externally, the distance between their centres equals—

Circle — সব প্রশ্ন দেখুন (231টি) →

In the proof of Theorem 26, △PAO\triangle PAO△PAO and △PBO\triangle PBO△PBO are—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof of Theorem 26, $\triangle PAO$ and $\triangle PBO$ are—