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Answer 1

11. Option d
12. Distance formula =root (x1-x2)^2 +(y1-y2)^2
Ans is option a
13.option b
14.option a
15 option d
16 option b
17 option b
18 option a
19 option d
20. Option a

Answer 2

$\sf\huge\bold{Answer:}$

11.

→ cos 60° = x

Value of cos 60° = ½

→ ½ = x

Hence, value of x is ½.

$\fbox{\bf\red{option\:d)\:is\:correct}}$

12.

Given : A(0,8) & B(0,-6)

To find : d(A, B)

Solution : Using distance formula

$\boxed{\tt d = \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} } }$

${\tt AB = \sqrt{(0 - 0) {}^{2} + (-6 -8) {}^{2} } }$

${\tt AB = \sqrt{(0 ) {}^{2} + (-14) {}^{2} } }$

${\tt AB = \sqrt{0 + 196 } }$

${\tt AB = \sqrt{196 } }$

${\tt \purple{AB = 14\:units }}$

d(A, B) = 14 units

$\fbox{\bf\red{option\:a)\:is\:correct}}$

13.

To find : Value of tanθ+cotθ

Solution : tanθ+cotθ

we know,

• tanθ=sinθ/cosθ
• cotθ=cosθ/sinθ

= tanθ+cotθ

= sinθ/cosθ + cosθ/sinθ

Taking LCM

= sin²θ+cos²θ/sinθcosθ

Using identity : sin²θ+cos²θ = 1

= 1/sinθcosθ

• 1/sinθ=cosecθ
• 1/cosθ=secθ

= (1/sinθ)×(1/cosθ)

= (cosecθ)×(secθ)

= cosecθsecθ

= secθcosecθ

$\fbox{\bf\red{option\:b)\:is\:correct}}$

14.

Given: XY || BC

Solution :

XY || BC & AB is transversal

∠AXY=∠ABC[corresponding angles]

XY || BC & AC is transversal

∠AYX = ∠ACB [corresponding angles]

In ∆AXY & ∆ABC

∠AXY=∠ABC[corresponding angles]

∠AYX = ∠ACB [corresponding angles]

∠XAY=∠BAC [common angle]

By AAA similarity criteria.

∆AXY ∼ ∆ABC

Similarly, ∆AYX∼ ∆ACB [as corresponding angles are equal to each other.]

$\fbox{\bf\red{option\:a)\:is\:correct}}$

15.

Given:QR = 6√3

∠Q=90°

∠R=30°

∠P=60°

To find : PR

Solution:

In ∆PQR,

QR is a base & PR is hypotenuse.

Also,

For θ = 30°

→cosθ= adjacent side(base)/hypotenuse

→cos 30° = QR/PR

→√3/2 = 6√3/PR

→PR = 6√3/(√3/2)

→PR = (6√3 × 2)/√3

→ PR = 12√3/√3

→ PR = 12

$\fbox{\bf\red{option\:d)\:is\:correct}}$

16.

m(arc AXB) = 100°

∠AOB = 100°

∠APB=½∠AOB

∠APB=½(100°)

∠APB=50°

i.e. m∠APB=50°

$\fbox{\bf\red{option\:b)\:is\:correct}}$

17.

• Four common tangents can be drawn to two disjoint circles.
• The tangents can be either direct or transverse.

Refer attachment for figure

$\fbox{\bf\red{option\:b)\:is\:correct}}$

18.

When a line is divided into two sections

First section : Second section = m:n

Point 1 of line segment(x1,y1)

Point 2 of line segment(x2,y2)

Section Formula

x = mx2+nx1/m+n, y = my2+ny1/m+n

$\fbox{\bf\red{option\:a)\:is\:correct}}$

19.

sinθ=7/25

cosθ=?

we know, sinθ = opposite side(perpendicular)/hypotenuse

& cosθ=adjacent side(base)/hypotenuse

By Pythagoras property

H²=B²+P²

25²=B²+7²

625=B²+49

B²=625-49

B²=576

B=√576

B=24

→cosθ=adjacent side(base)/hypotenuse

→cosθ=24/25

$\fbox{\bf\red{option\:d)\:is\:correct}}$

20.

∆1 ∼∆2

A(∆1)=3A(∆2)

S1 : S2 =?

When two triangles are similar, square of sides in ratio is equal to ratio of their areas

ar(∆1) / ar(∆2) = (S1/S2)²

3A(∆2) / A(∆2) = (S1/S2)²

3(∆2) / 1(∆2) = (S1/S2)²

3 / 1 = (S1/S2)²

(S1/S2)² = 3/1

(S1/S2) = √(3/1)

(S1/S2) = √3 /√1

(S1/S2) = √3 /1

S1:S2 = √3 : 1

$\fbox{\bf\red{option\:c)\:is\:correct}}$