**CAT Questions On Lines And Angles PDF:**

Download CAT Questions On Lines And Angles PDF. Practice important problemsĀ on CAT geometry Questions On Lines And Angles with detailed explanations and solutions.

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**Question 1:** In a triangle ABC, Angle B = 60 degres and Angle C = 45 degrees. D is a point on BC such that AD divides BC in the ratio 1:4. Find the value of $\left \lfloor \frac{sin(BAD)}{sin(CAD)} \right \rfloor$.$\left \lfloor \ \right \rfloor$ refers to the greatest integer function less than or equal to x.

a) 1

b) 0

c) 3

d) 5

**Question 2:**Ā A regular polygon P has 135 diagonals. Find the exterior angle of the polygon P.

a)Ā $18^{\circ}$

b)Ā $20^{\circ}$

c)Ā $25^{\circ}$

d)Ā $30^{\circ}$

**Question 3:**Ā A circle touches all the three sides of the triangle as shown in the given figure at points P, Q and R respectively. If Angle RPQ is 50 degrees, find the measure of angle BAC (in degrees)?

a) 65

b) 70

c) 80

d) 96

**Question 4:**Ā Tangents are drawn from the points (6,8) to the circle $x^2+y^2=36$. Find the angle between the two tangents approximately.

a)Ā $64^{\circ}$

b)Ā $74^{\circ}$

c) $84^{\circ}$

d)Ā $94^{\circ}$

**Question 5:**Ā A square ABCD has a side of length 3cm. A point E on CD divides CD in such a way that CE:ED = 1:2. AC and BE intersect at P. Find cos(Angle EPC).

a)Ā $\frac{1}{\sqrt{20}}$

b)Ā $\frac{1}{\sqrt{10}}$

c)Ā $\frac{2}{\sqrt{10}}$

d)Ā $\frac{1}{\sqrt{5}}$

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__Answers & Solutions:__

**1) Answer (B)**

In triangle ABD,

$\frac{sin(BAD)}{x} =\frac{sin 60}{L}$

In triangle CAD,

$\frac{sin(CAD)}{4x} =\frac{sin 45}{L}$

$4\frac{sin(BAD)}{sin(CAD)} = \frac{\sqrt{3}}{\sqrt{2}} = 1.22$

=> $\frac{sin(BAD)}{sin(CAD)} = 1.22/4 = .305$

$\left \lfloor \frac{sin(BAD)}{sin(CAD)} \right \rfloor$ = $\left \lfloor .305 \right \rfloor$ = 0

**2) Answer (B)**

Number of diagonals in a regular polygon = $^nC_2 – n$

=> $^nC_2 – n$ = 135

=> n = 18

Exterior angle = $\frac{360}{n}$ = $\frac{360}{18}$ = $20^{\circ}$

**3) Answer (C)**

Angle = ROQ = 2*50 = 100 degrees

Since OR = OQ

Angle ORQ = Angle OQR = (180-100)/2 = 40 degrees

Since line drawn from the centre to the point of contact is perpendicular to the tangent,

Angle ARQ = Angle AQR = 90-40 = 50 degrees

Angle RAQ = 180-50-50 = 80 degrees

**4) Answer (B)**

Let the angle between the tangents be 2x.

The radius of the circle is 6 and the center is (0,0).

=> Distance betweenthe center and point (6,8) is $\sqrt{6^2+8^2}$ = 10

=> Length of the tangent = $\sqrt{10^2-6^2}$ = 8

$\sin x$ = 6/10 = 0.6

$\sin 30^{\circ}$ = 0.5 and $\sin 45^{\circ}$ = $\frac{1}{\sqrt{2}}$ = 0.7

=> x must be around $35^{\circ}$ to $38^{\circ}$

=> 2x must be around $70^{\circ}$ to $76^{\circ}$

Among the options, option B is the closest answer.

**5) Answer (D)**

The diagram is as shown below.

Angle PCE = 45 degrees

Angle PEC =$cos^{-1}$(1/$\sqrt{10}$)

Angle EPC = 180-45-$cos^{-1}$(1/$\sqrt{10}$) = 135-$cos^{-1}$(1/$\sqrt{10}$)

cos(Angle EPC) = cos(135-$cos^{-1}$(1/$\sqrt{10})$ ) = cos(135)cos($cos^{-1}$(1/$\sqrt{10}$))+sin135sin($cos^{-1}$(1/$\sqrt{10})$)

cos(Angle EPC) = -$1/\sqrt{20}$+$3/\sqrt{20}$ = $\frac{1}{\sqrt{5}}$