Trigonometry & Height and Distances
Question 1 |
What is the maximum value of 5 Sinθ + 12 cosθ?
16 | |
13 | |
10 | |
12 |
Discuss it
Question 1 Explanation:
Maximum value = √(a2 + b2)
=13
Question 2 |
What is minimum value of Sinθ + cosθ ?
-2 | |
1/2 | |
-1 | |
-√2 |
Discuss it
Question 2 Explanation:
Minimum value = √(a2 - b2)
=-√2
=-√2
Question 3 |
If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?
2/√3 | |
1/2 | |
1/√2 | |
1/√3 |
Discuss it
Question 3 Explanation:
tanA = cotB,
tanA*tan B = 1
So, A +B = 90o
(x+y)+(x-y) = 90o, 2x = 90o , x = 45o
Tan (2x/3) = tan 30o = 1/√3
tanA*tan B = 1
So, A +B = 90o
(x+y)+(x-y) = 90o, 2x = 90o , x = 45o
Tan (2x/3) = tan 30o = 1/√3
Question 4 |
Find the Value of tan60o + tan120o?
1/2 | |
0 | |
2 | |
2/3 |
Discuss it
Question 4 Explanation:
=tan(60) + tan(180-120)
=tan(60)+(-tan(60)) {because in second quadrant tan is -ve}
=0
Question 5 |
The least value of 2sin2θ + 3cos2θ
1/3 | |
4/3 | |
2 | |
3/4 |
Discuss it
Question 5 Explanation:
= 2sin2θ + 2cos2θ + cos2θ
=2(sin2θ + cos2θ) + cos2θ ; (by putting sin2θ + cos2=1)
= 2 + cos2θ ;(the minimum value of cos2θ=0) = 2 + 0 = 2
=2(sin2θ + cos2θ) + cos2θ ; (by putting sin2θ + cos2=1)
= 2 + cos2θ ;(the minimum value of cos2θ=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:
30 degrees | |
45 degrees | |
60 degrees | |
90 degrees |
Discuss it
Question 6 Explanation:
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:
149 m | |
156 m | |
173 m | |
200 m |
Discuss it
Question 7 Explanation:
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?
4(3)^0.5 units | |
8 units | |
12 units | |
Data inadequate |
Discuss it
Question 8 Explanation:
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?
32 kmph | |
36 kmph | |
40 kmph | |
44 kmph |
Discuss it
Question 9 Explanation:
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?
πd | |
πd2 | |
(1/4) πd2 | |
(1/2)πd |
Discuss it
Question 10 Explanation:
One important observation to solve the question : 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
There are 11 questions to complete.
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