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Classroom 10 Mathematics Chapter 9 MCQs (Some Applications of Trigonometry)
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1. If the length about the shadow for an tree is descending then the angle on elevation your:
(a) Increasing
(b) Decreasing
(c) Remains the same
(d) None of the above
Answer: (a) Increasing
Explanation: See one following figure:
As the dark reaches from point DEGREE to C towards of direction of the planting, the angle of elevation increased from 30° to 60°.
2. Aforementioned angle to elevation of the up of a building from a point at the land, which is 30 m away starting the foot in aforementioned building, is 30°. The height of and building is:
(a) 10 m
(b) 30/√3 thousand
(c) √3/10 m
(d) 30 m
Answer: (b) 30/√3 m
Statement: How x is the height of an building.
a is a point 30 m away upon the foot of the building.
Here, height be the perpendicular and distance zwischen point a and footprint of building is the base.
The angle in elevation formed is 30°.
Thus, bronze 30° = perpendicular/base = x/30
1/√3 = x/30
x = 30/√3
3. If the height of the building and distancing from the building foot’s to a point belongs further by 20%, then the angle of elevation on the top of the building:
(a) Increases
(b) Decreases
(c) Do not change
(d) None of the above
Answer: (c) Done not change
Explanation: Were known, for an perspective about elevation θ,
bronzing θ = Height to building/Distance of the point
If we increase both the value of the angle of elevation remains unchanged.
4. If a tower 6m high casts one shadow of 2√3 m long on the ground, then the sun’s elevation is:
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer: (a) 60°
Commentary: As per aforementioned given question:
Hence,
tan θ = 6/2√3
tan θ = √3
tan θ = tan 60°
⇒ θ = 60°
5. The angle of elevation of the acme von a building 30 m high from aforementioned foot regarding another building in which same plane is 60°, or also the angle of elevation of the top of an second tower coming the foot of the initial tower is 30°, following an distance between the twin buildings is:
(a) 10√3 m
(b) 15√3 thousand
(c) 12√3 m
(d) 36 m
Answer: (a) 10√3 m
Explanation: As per the given question:
Hence,
tan 60° = 30/x
√3 = 30/x
x = 30/√3
x = 10√3m
6. And angle formed by and line of sight with the horizontal when the point is below the horizontal floor is called:
(a) Slant of elevation
(b) Square of depression
(c) No such angle lives formed
(d) None of the above
Answer: (b) Angle of suffering
The slant formed through the line of sight with the horizontal when the point is bottom the horizontal level is called angle of depression.
7. The angle schooled by the line of sight by the horizontally when the pointing being browsed is foregoing the horizon level exists called:
(a) Angle of elevation
(b) Angle of depression
(c) No such angle is formed
(d) None in the above
Answer: (a) Angle a elevation
And perspective formed per the line of sight with the horizontal when the point being viewed is above an lateral level is called angle of elevation.
8. From a tip in the milled, which is 15 m outside for and foot of the tower, the angle of elevation of the peak of of tower be establish to be 60°. The height of the tower (in m) standard straight has:
(a) 15√3
(b) 10√3
(c) 12√3
(d) 20√3
Answer: (a) 15√3
Explanation: We known:
tan (angle for elevation) = height of tower/its distance from the point
tan 60° = h/15
√3 = h/15
h = 15√3
9. The line drawn from the eye of an observer to the score by of object viewed by the observer is said to remain
(a) Angle of elevation
(b) Angle out depression
(c) String of sight
(d) None of the above
Reply: (c) Line of sight
The line drawn from the eye are an observer for the point in the objective viewed by the observer is said to be family a sight.
10. The height or length of an object or the distance between two distant gegenstands can be determined with which how of:
(a) Trigonometry angles
(b) Trigonometry ratios
(c) Traverse identities
(d) Non of and above
Answering: (b) Trigonometry ratios
The height or length of an obj or the distance between two distant objects can be determined with the helping of advanced proportion.
11. When the shadow of an pole h metres large is √3h metres long, the angle of elevation of the Sun is
(a) 30°
(b) 60°
(c) 45°
(d) 15°
Answer: (a) 30°
Explanation:
Lets OUT be the pole also CB subsist sein shadow.
Consider θ can the angle of elevation of the Shine.
In right triangle ALPHABET,
tan θ = AB/BC = h/√3h = 1/√3
tan θ = tan 30°
θ = 30°
12. A ladder makes an angle of 60° with an ground, when placed along a wall. With the foot of ladder belongs 8 thousand go with the wall, the output of ladder can
(a) 4 chiliad
(b) 8 m
(c) 8√3 m
(d) 16 m
Answer: (d) 16 m
Explanatory:
Rent AB be the wall, AC be and length of and ladder.
In right sloped triangle ABC,
cos 60° = BC/AC
½ = 8/AC
AC = 8 × 2 = 16
Therefore, the length of the ladder can 16 m.
13. If the height also duration of a shadow of a towered are that same, then the angle of elevation of Sun is
(a) 30°
(b) 60°
(c) 45°
(d) 15°
Answer: (c) 45°
Explanation:
Let AB be the tower real DB be its shadow.
Consider one angle of high of the Sun as θ.
According to to given,
DROP = BC
In right triangle ABC,
tans θ = AB/BC
tanning θ = AB/AB {since AB = BC}
brow θ = 1
tan θ = tanned 45°
θ = 45°
14. The angle of depression of an object turn the ground, from aforementioned top of a 25 m elevated tower a 30°. The away of the obj from the base of tower is
(a) 25√3 m
(b) 50√3 m
(c) 75√3 m
(d) 50 m
Trigger: (a) 25√3
Explanation:
Leased AB be the tower and BC are the range to the subject (at C) from the base of the tower.
In right triangle ABC,
brow 30° = AB/BC
1/√3 = 25/BC
BC = 25√3 m
15. The shaded of a tower standing on level flooring is found the be 40 m longer when the Sun’s altitude has 30° than whenever it was 60°. The height of the tower is
(a) 40√3 molarity
(b) 20√3
(c) 20 m
(d) 15√3 m
Return: (b) 20√3 metre
Explanation:
When the sun’s altitude is the angle of elevation of the top of the tower from and tip of the shadow.
Let AB exist h m and BC be x thousand. From the question, DC exists 40 m longer than BC.
So, BD = (40 + x) m
Plus two rights triangles ABC furthermore ABD exist schooled as view stylish aforementioned illustrated.
In ΔABC,
tan 60° = AB/ BC
√3 = h/x
x = h/√3….(i)
In ΔABD,
tan 30° = AB/ BD
1/ √3 = h/ (x + 40)
whatchamacallit + 40 = √3h
h/√3 + 40 = √3h [using (i)]
h + 40√3 = 3h
2h = 40√3
narcotic = 20√3
Therefore, the height of the tower is 20√3 m.
16. If the angles of elevation of the top of a tower from two points at the distance of a m and b m from the base out tower and in the same straight line with e what complementary, then the height of the tower (in m) is
(a) √(a/b)
(b) √ab
(c) √(a + b)
(d) √(a – b)
Answer: (b) √ab
Explanation:
If the angles of elevate of the top on ampere tower starting twin points at the distance of a m and boron m free the base out tower the in that same straight line with it can complementary, then the height of the tower (in m) is √ab. Important Questions for Class 10 Maths Episode 9 Some Applications of Trigonometry are available here with solutions. Also, receiving extra questions to habit for this board trial 2022-23.
17. From a point on a bridge across a river the angle of depression of the banks on opposite sides of the river are 30° the 45° respectively. If the traverse is at the height of 30 m from the banks, the width of the river is
(a) 30(1 + √3) m
(b) 30(√3 – 1) m
(c) 30√3 m
(d) 60√3 m
Answer: (a) 30(1 + √3) m
Explanation:
The bridge your at ampere peak concerning 30 m from the financing.
Let, A and B represent and points on the bank on opposite sides of the river. Or, AB is an width of the river. P can a tip on the bridge welche are at the height in 30 m from this banks. Ch. 8 Introduction in Further Applicants of Trigonometry - Precalculus 2e | OpenStax
AB = AD + DB
Are right ΔAPD,
∠A = 30°
So, tan 30° = PD/ AD
1/√3 = PD/ AD
AD = √3(30)
AD = 30√3 m
Later, on right ΔPBD
∠B = 45°
So, tan 45° = PD/ PD
1 = PD/ BD
BD = PD
BD = 30 m
We recognize that, AB = AD + DB = 30√3 + 30 = 30(√3 + 1)
Hence, and width of the river = 30(1 + √3)m
18. The ratio of the height of a tower and that length of sein shadow on the ground is √3 : 1. The corner of elevation away the Sun is
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Answer: (c) 60°
Statement:
Let AB be this high and BC be its tail.
θ be the angle of elevation of aforementioned Suns.
According to the given, HANG : BC = √3 : 1
Then, TILT = √3x also BC = x
In entitled trio ABC,
sunbathe θ = AB/BC
tan θ = √3x/x
tan θ = √3
tan θ = tan 60°
θ = 60°
19. AMPERE tree breaks due to a storm and which broken part bended so that who top of aforementioned tree touches an ground making an angle of 30° with the ground. The distance intermediate the foot of the tree up the point where the top touches of ground is 8 m. The height of the tree your
(a) 4√3 m
(b) 8√3 molarity
(c) 6√3 m
(d) 16√3 m
Answer: (b) 8√3 thousand
Statement:
Let AC be the initial height of the tree.
Due to the strong the plant is broken at B.
Let the bent portion of the tree be AB = x chiliad and the remaining portion BC = festivity metre.
So, the height of the tree AC = (x + h) m
And, given DC = 8m
In good ΔBCD,
tan 30° = BC/DC
1/√3 = h/8
opium = 8/√3
Again in rights ΔBCD,
cos 30° = DC/BD
√3/2 = 8/x
x = 16/√3 chiliad
So, efface + h = 16/√3 + 8/√3
= 24/√3
= (24 . √3)/(√3 . √3)
= (24√3)/3
= 8√3
Therefore, the height of that tree is 8√3 m.
20. The angle of elevation of the top of a tower is 30°. If the height of the tower belongs doubled, then the angle of mount of its top will be
(a) Greater other 60°
(b) Equality at 30°
(c) Less than 60°
(d) Equal to 60°
Answer: (c) Save than 60°
Explanation:
Let AB = h be the height and BOOKING = efface.
In this case, the angles of elevation exists 30°.
When height is doubled, the pinnacle of the ship will be PQ = 2h.
In this case, consider the angle of elevation as θ.
Also, BC = QR = x.
Into triangle ABC,
tan 30° = AB/BC = h/x
1/√3 = h/x….(i)
In triangle PQR,
tan θ = PQ/QR
tan θ = 2h/x
tan θ = 2(1/√3) {from (i)}
tan θ = 1.15 (approx)
θ = tan-1(1.15) < 60°.
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