Question 11: In following figure, the angles of depressions from the observing positions O_{1} and O_{2} respectively of the object are ……., ……………..

**Answer:** O_{1} 30° and O_{2} 45°

Question 12: In &Detla; ABC, Ab = `6sqrt3` cm, AC = 12 cm and BC = 6 cm, then ∠B = ?

Answer: 90°

OR

Two triangles are similar if their corresponding sides are ………………

**Answer:** In the same ratio

Question 13: In following figure, the length of PB = ?

**Answer:** 4 cm (Note: 3, 4 and 5 make Pythagorean triplet)

Question 14: In following figure, MN || BC and AM : MB = 1 : 2, then

`text(arΔAMN)/text(arΔABC)` = ?

**Answer:** 1 : 9 (Ratio of areas is square of ratio of sides)

Question 15: The value of sin 32° cos 58° + cos 32° sin 58° is …………..

**Answer:** 1

**Explanation:** sin 32° cos 58° + cos 32° sin 58°

= sin 32° × cos(90°-32°) + cos 32° × sin(90°-32°)

= sin^{2} 32° + cos^{2} 32° = 1

Question 16: A die is thrown once. What is the probability of getting a prime number?

**Answer:** P_{(E)} = `3/6=1/2`

There are three primer numbers below 6, i.e. 2, 3 and 5

Question 17: If a number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3, then find the probability of of `x^2 &let; 4`

**Answer:** `3/7`

OR

What is the probability that a randomly taken leap year has 52 Sundays?

**Answer:** `5/7`

**Explanation:** There are 366 days in a leap year. So, a leap year has 52 weeks and 2 days. These extra 2 days can be any of the following combinations:

**Sunday-Monday**, Monday-Tuesday, Tuseday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, **Saturday-Sunday**

So, out of 7 possible combinations, two have Sundays which means there will be more than 52 Sundays on these two occasions. Rest of the occasions will have exactly 52 Sundays.

Question 18: If sin A + sin^{2} = 1, then find the value of the expression (cos^{2}A + cos^{4}A).

**Answer:** 1

**Explanation:** sin A + sin^{2} = 1

Or, sin^{2}A = 1 – sin A ……………(1)

Now, cos^{2}A + cos^{4}A

= 1 – sin^{2}A + (cos^{2}A)^{2}

= 1 – sin^{2}A + (1 – sin^{2}A)^{2}

Substituting the value of sin^{2}A from equation 1, we get:

1 – (1 – sin A) + (1 – (1 – sin A)^{2}

= 1 – 1 + sin A + (1 – 1 + sin A )^{2}

= sin A + sin^{2}A = 1

Question 19: Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π = 3.14).

**Answer:** Area of sector `=πr^2×θ/(360°)`

`=3.14×6^2×(30°)/(360°)`

`=3.14×36×1/(12)=3.14×3=9.42` sq cm

Question 20: Find the class marks of the classes 20 – 50 and 35 – 60.

**Answer:** 35 and 47.5

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