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QUESTION: 1

What is the maximum value of 5 Sinθ + 12 cosθ?

Solution:

Maximum value = √(a^{2} + b^{2})

=13

QUESTION: 2

What is minimum value of Sinθ + cosθ ?

Solution:

Minimum value = √(a^{2} – b^{2})

=-√2

QUESTION: 3

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

Solution:

tanA = cotB,

tanA*tan B = 1

So, A +B = 90^{o}

(x+y)+(x-y) = 90^{o}, 2x = 90^{o} , x = 45^{o}

Tan (2x/3) = tan 30^{o} = 1/√3

QUESTION: 4

Find the Value of tan60^{o} + tan120^{o}?

Solution:

tan(60) + tan(180-120)

=tan(60)+(-tan(60)) {because in second quadrant tan is -ve}

=0

QUESTION: 5

The least value of 2sin^{2}θ + 3cos^{2}θ

Solution:

2sin^{2}θ + 2cos^{2}θ + cos^{2}θ

=2(sin^{2}θ + cos^{2}θ) + cos^{2}θ ; (by putting sin^{2}θ + cos^{2}=1)

= 2 + cos^{2}θ ;(the minimum value of cos^{2}θ=0)

= 2 + 0 = 2

QUESTION: 6

The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree, is:

Solution:

Let AB be the height of the tree and AC be the length of the shadow. We need to calculate the angle ACB where BC is the hypotenuse.

Given,

AC:AB = √3

cot θ = √3

θ = 30 degrees.

QUESTION: 7

From a point P on a level ground, the angle of elevation of the top of a tower is 30 degrees. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Solution:

Let QR be the tower. Then,

QR = 100 m and angle QPR = 30 degrees.

We know, cot 30° = √3 = PQ/QR.

Therefore, PQ = 100*√3 = 173 m.

QUESTION: 8

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of elevation becomes 60°. What is the distance between the base of the tower and the point P?

Solution:

Let MN be the tower and man be standing at P (30° = angle MPN) and Q (60° = angle MQN). We are only given two angles and no sides of the triangles. Therefore, the data is inadequate.

QUESTION: 9

A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 60 meters from the tower. After 5 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?

Solution:

Let the tower be PQ and the boat be at positions R and S when making angles of 45° and 30° respectively.

Given, PR = 60 m.

Now, PQ/PR = tan 45° = 1. So, PQ = PR = 60 m.

Again, PQ/PS = tan 30° = 1/√3. So, PS = 60 * √3 m = 103.92 m.

Distance covered in 5 seconds = 103.92 – 60 = 43.92 m.

Speed in kmph = (43.92/5) * (18/5) = 32 kmph (approximately)

QUESTION: 10

The area of a square is ‘d’. What is the area of the circle which has the diagonal of the square as its diameter?

Solution:

Diagonal of Square = Diameter of Circle.

Let side of square be x.

From Pythogorous theorem.

Diagonal = √(2*x*x)

We know area of square = x * x = d ⇒ Diameter = Diagonal = √(2*d)

⇒ Radius = √(d/2)

Area of Circle = π * √(d/2) * √(d/2)

= 1/2 * π * d

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