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AP Polycet (Polytechnic) 2022 Previous Question Paper with Answers And Model Papers With Complete Analysis

31) If a point P is 17cm from the center of a circle of radius 8cm, then the length of the tangent drawn to the circle from the point P is

A) 10cm

B) 12cm

C) 15cm

D) 14cm

View Answer

C) 15cm
Explanation:Given: Distance from point P to center = 17 cm, Radius of circle = 8 cm
We use the Pythagorean theorem for the right triangle formed by radius, tangent, and line from P to center.
Let tangent length = t
$t^2 = 17^2 - 8^2 = 289 - 64 = 225 ⇒ t = \sqrt{225} = \boxed{15 \text{ cm}}$

32) If cos A= $\frac45$ , then the value of tan A is

A) $\frac35$

B) $\frac34$

C) $\frac43$

D) $\frac53$

View Answer

B) $\frac34$
Explanation:Given: $\cos A = \frac{4}{5}$
Use identity: $\sin^2 A + \cos^2 A = 1 ⇒ \sin A = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$
$\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \boxed{\frac{3}{4}}$

33) The value of $\frac{cot45^\circ}{\sin\left(30^\circ\right)+\cos\left(60^\circ\right)}$ is equal to

A) 1

B) $\frac1{\sqrt2}$

C) $\frac23$

D) $\frac12$

View Answer

A) 1
Explanation:
$\cot 45^\circ = 1, \quad \sin 30^\circ = \frac{1}{2}, \quad \cos 60^\circ = \frac{1}{2}$
$\frac{\cot 45^\circ}{\sin 30^\circ + \cos 60^\circ} = \frac{1}{\frac{1}{2} + \frac{1}{2}} = \frac{1}{1} = \boxed{1}$

34) The value of tan 2°. tan 4°.tan 6°…tan 88° is

A) 0

B) 1

C) 2

D) Not defined

View Answer

B) 1
Explanation:This is a known trigonometric product:
$\tan 2^\circ \cdot \tan 4^\circ \cdot \tan 6^\circ \cdots \tan 88^\circ = \boxed{1}$

35) If tanθ+cotθ=5, then tan²θ+cot²θ=?

A) 27

B) 25

C) 24

D) 23

View Answer

D) 23
Explanation:Given: $\tan \theta + \cot \theta = 5$
Use identity:
$(\tan \theta + \cot \theta)^2 = \tan^2 \theta + \cot^2 \theta + 2 ⇒ 25 = \tan^2 \theta + \cot^2 \theta + 2 ⇒ \tan^2 \theta + \cot^2 \theta = 25 - 2 = \boxed{23}$

36) If x=asinθ and y=btanθ, then the value of $\frac{a^2}{x^2}-\frac{b^2}{y^2}$ is

A) 1

B) 2

C) -1

D) None of these

View Answer

A) 1
Explanation:Given: $x = a \sin \theta, y = b \tan \theta$
Then:
$\frac{a^2}{x^2} = \frac{1}{\sin^2 \theta}, \quad \frac{b^2}{y^2} = \frac{1}{\tan^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} ⇒ \frac{a^2}{x^2} - \frac{b^2}{y^2} = \frac{1 - \cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = \boxed{1}$

37) If secθ+tanθ=x, then tan θ=

A) $\frac{x^2+1}x$

B) $\frac{x^2-1}x$

C) $\frac{x^2+1}{2x}$

D) $\frac{x^2-1}{2x}$

View Answer

D) $\frac{x^2-1}{2x}$
Explanation:Given: $\sec \theta + \tan \theta = x$
Use identity:
$(\sec \theta + \tan \theta)^2 = \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta = x^2 ⇒ 1 + \tan^2 \theta + \tan^2 \theta + 2 \tan \theta \sqrt{1 + \tan^2 \theta} = x^2$
Too long—use shortcut:
From identity:
$\tan \theta = \frac{x^2 - 1}{2x}$

38) $\frac{\sin\left(\theta\right)}{1+\cos\left(\theta\right)}$ =

A) $\frac{1+\cos\left(\theta\right)}{\sin\left(\theta\right)}$

B) $\frac{1-\cos\left(\theta\right)}{\cos\left(\theta\right)}$

C) $\frac{1-\cos\left(\theta\right)}{\sin\left(\theta\right)}$

D) $\frac{1+\sin\left(\theta\right)}{\cos\left(\theta\right)}$

View Answer

C) $\frac{1-\cos\left(\theta\right)}{\sin\left(\theta\right)}$
Explanation:Use identity:
$\frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta} \quad \text{(rationalizing)}$
Answer: $\frac{1 - \cos \theta}{\sin \theta}$

39) If the ratio of the length of a pole and its shadow is 1:√3, then the angle of elevation of the

A) 30°

B) 45°

C) 60°

D) 90°

View Answer

C) 60°
Explanation:Pole\:Shadow = 1:√3 → Use basic trigonometry:
$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} ⇒ \theta = \boxed{30^\circ}$

40) If two towers of heights h₁ and h₂ subtend angles 45° and 30° respectively at the midpoint of the line joining their feet, then the ratio of h₁: h₂ is

A) 1:√3

B) √3:1

C) 1:3

D) 3:1

View Answer

B) √3:1
Explanation:Use:
$\tan 45^\circ = \frac{h_1}{d}, \quad \tan 30^\circ = \frac{h_2}{d} ⇒ h_1 = d, h_2 = d \cdot \tan 30^\circ = d \cdot \frac{1}{\sqrt{3}} ⇒ \frac{h_1}{h_2} = \frac{d}{d/\sqrt{3}} = \boxed{\sqrt{3}:1}$

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