Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then is equal

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.
Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then ${(Q R)}^{2}$ is equal to _________. [2025]

Ans.
(28)

Let $∠ P R N = θ$

$\Rightarrow P R = \cos e c θ and P Q = 4 \sec (30 ° + θ)$

For equilateral $△$ PQR.

Let PR = PQ

$\Rightarrow cosec θ = 4 \sec (θ + 30 °)$

$\Rightarrow \cos (θ + 30 °) = 4 \sin θ$

$\Rightarrow \frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ = 4 \sin θ$

$\Rightarrow \tan θ = \frac{1}{3 \sqrt{3}}$

$\Rightarrow cosec θ = \sqrt{28}$

Now, $Q R^{2} = d^{2} = {cosec}^{2} θ = 28$ .

Want Us to Email you About Special Offers & Updates?