In the given figure, PQ is tangent to a circle at O and , show that BP = BQ.

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
In the given figure, PQ is tangent to a circle at O and $∠ B A Q = 30 °$ , show that BP = BQ.

Ans.
$∠ B Q P = ∠ B A Q (∠ s in alternate segment are equal)$

$\Rightarrow ∠ B Q P = 30 ° ...(i) (∵ ∠ B A Q = 30 ° given)$

As AB is a diameter, AQB is a Semicircle.

$∠ A Q B = 90 ° (angle in semicircle = 90 °)$

From figure, $∠ A Q P = ∠ A Q B + ∠ B Q P$

$\Rightarrow ∠ A Q P = 90 ° + 30 ° = 120 °$

In $∆ A Q P,$ $∠ Q P A + ∠ B A Q + ∠ A Q P = 180 °$

$\Rightarrow ∠ Q P A + 30 ° + 120 ° = 180 °$

$\Rightarrow ∠ Q P A = 180 ° - (30 ° + 120 °)$

$\Rightarrow ∠ Q P A = 30 ° ...(ii)$

From (i) and (ii) we get

$∠ B Q P = ∠ Q P B = 30 °$

Therefore, QB = BP

Want Us to Email you About Special Offers & Updates?