In the given figure, the angles $\angle B A Q=\angle C P Q=\angle C B Q=\frac{\pi}{2}$; and the leng...

Let $\angle P Q R=\theta$,By angle sum property of triangles $\angle P R Q=90-\theta$Since opposite angles in intersecting lines are equal $\angle B R C=90-\theta$Again, by using angle sum property of triangles, $\angle B C R=\theta$Now let us construct a perpendicular to $C R$ which intersect with point B which will be equal in length to $P A$Let us focus on $\triangle B S C$Where $B C=1$ unitHere,$$ \sin \theta=\frac{B S}{C B}=\frac{4}{5}(\text { from } \triangle A Q B) $$as in triangle $A Q B, B Q=5 \mathrm{~cm}$ (use Pythagorean theorem)So,$$ B S=\frac{4(1)}{5}=0.8 $$Since $P A=B S=0.8 \mathrm{~cm}$$$ Q P=Q A-P A=3-0.8=2.2 \text { unit } $$

Properties Of Triangles - IAT-IISER Maths PYQs

Let

%5Cangle%20P%20Q%20R%3D%5Ctheta

,By angle sum property of triangles

%5Cangle%20P%20R%20Q%3D90-%5Ctheta

Since opposite angles in intersecting lines are equal

%5Cangle%20B%20R%20C%3D90-%5Ctheta

Again, by using angle sum property of triangles,

%5Cangle%20B%20C%20R%3D%5Ctheta

Now let us construct a perpendicular to

C%20R

which intersect with point B which will be equal in length to

P%20A

Let us focus on

%5Ctriangle%20B%20S%20C

Where

B%20C%3D1

unitHere,$

%5Csin%20%5Ctheta%3D%5Cfrac%7BB%20S%7D%7BC%20B%7D%3D%5Cfrac%7B4%7D%7B5%7D(%5Ctext%20%7B%20from%20%7D%20%5Ctriangle%20A%20Q%20B)

as%20in%20triangle

A Q B, B Q=5 \mathrm{~cm}

(use%20Pythagorean%20theorem)So%2C

B%20S%3D%5Cfrac%7B4(1)%7D%7B5%7D%3D0.8

Since

P A=B S=0.8 \mathrm{~cm}$$

Q%20P%3DQ%20A-P%20A%3D3-0.8%3D2.2%20%5Ctext%20%7B%20unit%20%7D

Properties Of Triangles

High-Yield Trend

Questions
1 MCQs

Official Solution

Properties Of Triangles

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs