Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Apram Sachdeva 6 years, 1 month ago
- 4 answers
Posted by ~ Honey... ? 6 years, 1 month ago
- 16 answers
Lucifer ? Morningstar? 6 years, 1 month ago
Lucifer ? Morningstar? 6 years, 1 month ago
Lucifer ? Morningstar? 6 years, 1 month ago
Lucifer ? Morningstar? 6 years, 1 month ago
Lucifer ? Morningstar? 6 years, 1 month ago
Lucifer ? Morningstar? 6 years, 1 month ago
Posted by Pratap Ghanchi 6 years, 1 month ago
- 2 answers
Posted by Suhail Suhu 6 years, 1 month ago
- 3 answers
Posted by Angel Thankachan 6 years, 1 month ago
- 2 answers
Posted by Amal George 6 years, 1 month ago
- 1 answers
Posted by Ramila Devi 6 years, 1 month ago
- 0 answers
Posted by Shabana Parween 6 years, 1 month ago
- 1 answers
Jay Kumar Sharma 6 years, 1 month ago
Posted by Ayushi Kumari 6 years, 1 month ago
- 0 answers
Posted by Khushi Patidar 6 years, 1 month ago
- 1 answers
Posted by Jasleen Kaur 6 years, 1 month ago
- 0 answers
Posted by Abhay Kumar Yadav 6 years, 1 month ago
- 0 answers
Posted by Sahil Kumar 6 years, 1 month ago
- 0 answers
Posted by Mukesh Rawat 6 years, 1 month ago
- 2 answers
Panchali Pal 6 years, 1 month ago
Abhinav ? 6 years, 1 month ago
Posted by Pawan Kumar 6 years, 1 month ago
- 2 answers
Posted by Anitha P 6 years, 1 month ago
- 2 answers
Posted by Sakina Suwasrawala 6 years, 1 month ago
- 3 answers
Posted by Otaku Also Should Study 6 years, 1 month ago
- 1 answers
Sia ? 6 years, 1 month ago
We know that, the lengths of tangents drawn from an external point to a circle are equal.
{tex}\therefore
{/tex} TP = TQ
In {tex}\triangle{/tex}TPQ,
TP = TQ
{tex}\Rightarrow{/tex} {tex}\angle{/tex}TQP = {tex}\angle{/tex}TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)
{tex}\angle{/tex}TQP + {tex}\angle{/tex}TPQ + {tex}\angle{/tex}PTQ = 180º (Angle sum property)
{tex}\therefore
{/tex}2 {tex}\angle{/tex}TPQ + {tex}\angle{/tex}PTQ = 180º (Using(1))
{tex}\Rightarrow{/tex} {tex}\angle{/tex}PTQ = 180º – 2 {tex}\angle{/tex}TPQ ...(1)
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP {tex}\perp{/tex} PT,
{tex}\therefore
{/tex} {tex}\angle{/tex}OPT = 90º
{tex}\Rightarrow{/tex}{tex}\angle{/tex}OPQ + {tex}\angle{/tex}TPQ = 90º
{tex}\Rightarrow{/tex} {tex}\angle{/tex}OPQ = 90º – {tex}\angle{/tex}TPQ
{tex}\Rightarrow{/tex} 2{tex}\angle{/tex}OPQ = 2(90º –{tex}\angle{/tex}TPQ) = 180º – 2 {tex}\angle{/tex}TPQ ...(2)
From (1) and (2), we get
{tex}\angle{/tex}PTQ = 2{tex}\angle{/tex}OPQ
Posted by Chandan Kumar 6 years, 1 month ago
- 1 answers
Sia ? 6 years, 1 month ago
Since, P(E) + P (not E) = 1
P (not E) = 1 - P(E) = 1 - 0.05 = 0.95
Posted by Adil Rashid 6 years, 1 month ago
- 0 answers
Posted by Yangchen Sherpa 6 years, 1 month ago
- 3 answers
Sidharth ($Id) 6 years, 1 month ago
Posted by Dhruv Sharma 6 years, 1 month ago
- 3 answers
Posted by Amna Banu 6 years, 1 month ago
- 1 answers
Posted by Khushi Keshri 6 years, 1 month ago
- 0 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Shiva Prakash 6 years, 1 month ago
0Thank You