In the adjoining figure, T P and T Q are the tangents to the circle with center O. The measure of ? P T Q is, 5/29/2018 · In figure, if TP and TQ are the two tangents to a circle with centre O so that POQ = 110 , then PTQ is equal to 60 ( B ) 70 (C) 80 (D) 90 Given POQ = 110 Here TP is a tangent. So, OP TP Hence, OPT = 90 Similarly , TQ is a tangent. So, OQ TQ Hence, OQT = 90 Now, In quadrilateral TPOQ PTO + OQT + PTQ + POQ = 360 Putting values 90 + 90 + PTQ + 110 …
5/29/2018 · Example 2 Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ?PTQ = 2 ?OPQ Given: A circle with centre O Two tangents TP and TQ to the circle where P and Q are the point of contact To prove: ? PTQ = 2 ?OPQ Proof: We know from theorem 10.2 that length of tangents drawn from an external point to a circle are.
Home. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that angle angle POQ equals 110°, then angle PT Q is equal to. # NCERT.
We know that, the lengths of tangents drawn from an external point to a circle are equal. ? TP = TQ In ?TPQ, TP = TQ ? ?TQP = ?TPQ …(1) (In a triangle, equal sides have equal angles opposite to them) ?TQP + ?TPQ + ?PTQ = 180º (Angle sum property) ? 2 ?TPQ + ?PTQ = 180º (Using(1)) ? ?PTQ = 180º – 2 ?TPQ …(1), 10/30/2020 · In the adjoining figure , TP and TQ are the tangents to the circle with centre O. The measure of ?PTQ is (A) 90° (C) 70° (B) 110° (D) 40° Question 7. The coordinates of origin are (A) (1, 1) (B) (2, 2) (C) (0, 0) (D) (3, 3) Question 8. If the discriminant of quadratic equation b 2 – 4ac = 0 then the roots are (A) Real and distinct (B …
In the adjoining figure, TP and TQ are the two tangents to a circle with centre O. If +POQ =110c, then +PTQ is (a) 60c (b) 70c (c) 80c (d) 90c Ans : Here OP PT= and OQ QT= , In quadrilateral OPTQ, we have +POQO+++++PT PTQO QT = 360c 1109cc++09 +PTQ+0c =360c +PTQ =70c Thus (b) is correct option. 11. AB and CD are two common tangents to circles, Two tangent TP and TQ are drawn to a circle with center O from an external point T. Prove that angle PTQ = 2 angle OPQ 1 See answer … In the adjoining figure , O is the centre of the circle. If angle BCA = X -15° find measure o … f angle ABC – Y.Bes onthe perpendicular bisector from centre on chord AC.BC(a) 90°(b) 60°(C) 45°(d) 150° …
3/31/2018 · Observe the tangents to the circle in the figure given … The tangents at P and Q intersect at a point T (See figure ). Find the length of TP . Solution :- Join OT . Let OT intersect PQ at R From theorem 10.2, Length of tangents from external points are equal So, TP = TQ In /_ TPQ TP = TQ (i.e. two sides are equal) So, /_ TPQ is an isosceles …
In figure , if TP and TQ are the two tangents to a circle with centre O so that … In the adjoining figure , if AD, AE and BC are tangents to the circle at D, E and F respectively, then prove that 2 AD=AB+BC+CA. Ans. CD = CF, BE = BF. 32. In figure , PA and PB are tangents from P to the circle with centre O. R is a point on the circle,prove that …
