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1

WB JEE 2009

MCQ (Single Correct Answer)

In triangle ABC, a = 2, b = 3 and sin A = 2/3, thne B is equal to

A
30$$^\circ$$
B
60$$^\circ$$
C
90$$^\circ$$
D
120$$^\circ$$

Explanation

Apply $${{\sin A} \over a} = {{\sin B} \over b}$$ (sine Rule)

$${{2/3} \over 2} = {{\sin B} \over 3} \Rightarrow \sin B = 1 = \sin 90^\circ $$

$$ \Rightarrow B = 90^\circ $$

2

WB JEE 2009

MCQ (Single Correct Answer)

The smallest value of $$5\cos \theta + 12$$ is

A
5
B
12
C
7
D
17

Explanation

$$\because$$ $$ - 1 \le \cos \theta \le 1 \Rightarrow - 5 \le 5\cos \theta \le 5$$

$$ \Rightarrow 12 - 5 \le 5\cos \theta + 12 \le 5 + 12$$

$$ \Rightarrow 7 \le 5\cos \theta + 12 \le 17$$

$$\therefore$$ Smallest value of $$5\cos \theta + 12$$ is 7

3

WB JEE 2009

MCQ (Single Correct Answer)

A positive acute angle is divided into two parts whose tangents are 1/2 and 1/3. Then the angle is

A
$$\pi$$/4
B
$$\pi$$/5
C
$$\pi$$/3
D
$$\pi$$/6

Explanation

Let $$\alpha$$ + $$\beta$$ = $$\theta$$ where tan $$\alpha$$ = $${1 \over 2}$$, tan $$\beta$$ = $${1 \over 3}$$

$$\therefore$$ $$\tan \theta = \tan (\alpha + \beta ) = {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}$$

$$ = {{{1 \over 2} + {1 \over 3}} \over {1 - {1 \over 2}.{1 \over 3}}} = 1$$

$$ \Rightarrow \theta = {\pi \over 4}$$

4

WB JEE 2009

MCQ (Single Correct Answer)

$$P = {1 \over 2}{\sin ^2}\theta + {1 \over 3}{\cos ^2}\theta $$, then

A
$${1 \over 3} \le P \le {1 \over 2}$$
B
$$P \ge {1 \over 2}$$
C
$$2 \le P \le 3$$
D
$$ - {{\sqrt {13} } \over 6} \le P \le {{\sqrt {13} } \over 6}$$

Explanation

$$P = {1 \over 2}{\sin ^2}\theta + {1 \over 3}(1 - {\sin ^2}\theta ) = {1 \over 3} + {1 \over 6}{\sin ^2}\theta $$

$$0 \le {\sin ^2}\theta \le 1 \Rightarrow {1 \over 3} \le P \le {1 \over 2}$$

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