RBSE Class 10 Maths Important Questions Chapter 9 Some Applications of Trigonometry

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 9 Some Applications of Trigonometry Important Questions and Answers.

RBSE Class 10 Maths Chapter 9 Important Questions Some Applications of Trigonometry

Objective Type Questions—

Question 1.
If the shadow of a tree of height 10 m is 10 \(\sqrt {3}\) m, then the angle of elevation of the Sun is—
(A) 90°
(B) 60°
(C) 45°
(D) 30°
Answer:
(D) 30°

Question 2.
The shadow of a tower is equal to its height then the angle of elevation of the Sun is—
(A) 90°
(B) 60°
(C) 45°
(D) 30°
Answer:
(C) 45°

Question 3.
If the shadow of a tree is \(\frac{1}{\sqrt{3}}\) times its height, then the angle of elevation of the Sun is—
(A) 90°
(B) 60°
(C) 45°
(D) 30°
Answer:
(B) 60°

Question 4.
The slope of a hill makes an angle of 60° with the horizontal of one has to walk 500 m in reaching the top. Then the height of the hill is—
(A) 500 \(\sqrt {3}\) m
(B) \(\frac{500}{\sqrt{3}}\) m
(C) 250\(\sqrt {3}\) m
(D) \(\frac{250}{\sqrt{3}}\) m
Answer:
(C) 250\(\sqrt {3}\) m

Question 5.
If the height of a tower is \(\sqrt {3}\) times its shadow then the angle of elevation of the Sun is—
(A) 30°
(B) 45°
(C) 75°
(D) 60°
Answer:
(D) 60°

Question 6.
If the shadow of a tree 10 m high is 10\(\sqrt {3}\) m long then the angle of elevation of the Sun will be—
(A) 15°
(B) 30°
(C) 45°
(D) 60°
Answer:
(B) 30°

Question 7.
The angle of depression of a boat from the top of a 50 m high light house is 60°. The distance of the boat from the light house will be—
(A) 50\(\sqrt {3}\) m
(B) \(\frac{50}{\sqrt{3}}\) m
(C) 50 × 3 m
(D) 50 m
Answer:
(B) \(\frac{50}{\sqrt{3}}\) m

Very Short Answer Type Questions—

Question 1.
The shadow of a pillar 3 m high is 3 m. At that instant find the angle of elevation of the sun.
Solution:
tan θ = \(\frac{3}{3}\) = 1
∴ θ = 45°

Question 2.
In the given figure find the value of angle θ.

Solution:
tan θ = \(\frac{10}{10 \sqrt{3}}\)
= \(\frac{1}{\sqrt{3}}\)
∴ θ = 30°

Question 3.
Write the value of ‘x’ according to the figure—

AB = x,
BC = 6\(\sqrt {3}\) m
and ∠C = 30°
Solution:
tan 30° = \(\frac{x}{6 \sqrt{3}}\)
So x = 6\(\sqrt {3}\) × tan 30°
= 6\(\sqrt {3}\) × \(\frac{1}{\sqrt{3}}\)
= 6 मीटर

Question 4.
What change takes place in the length of the shadow of any pillar when angle of elevation of the Sun increases from 0° to 90°? Write.
Answer—
The length of the shadow will go on decreasing.

Question 5.
The angle of elevation of the top of a chimney from a distance of 200 m from the base of the chimney is 60°. Find the height of the chimney.
Solution:
tan 60° = \(\frac{x}{200}\) Or x = 200 × \(\sqrt {3}\)
So x = 200 × 1.732 = 346.4 m

Question 6.
From the bridge of a river 20 m high the angle of depression of a boat is 30°. Find how much has the boat to travel in reaching below the bridge?
Solution:
tan 30° = \(\frac{20}{x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{20}{x}\) [∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]
Or x = 20\(\sqrt {3}\) m

Question 7.
The height of a telephone pole from the ground is 17 m. The wire tied to its upper end makes an angle of 45° with the ground. Find the length of the wire.
Solution:
sin 45° = \(\frac{17}{x}\)
[∵ sin 45° = \(\frac{1}{\sqrt{2}}\)]
∴ \(\frac{1}{\sqrt{2}}\) = \(\frac{17}{x}\) Or x = 17\(\sqrt {2}\) = 17 × 1.414
= 24.04 m

Short Answer Type Questions—

Question 1.
A tree breaks at a height of 4 m from the ground and falls such that its upper end makes an angle of 30° from the ground. Find the total height of the tree.
Solution:
Here BC is the broken part of the tree.

Hence the total height of the tree (AC + BC) = 4 + 8 = 12 m

Question 2.
From the roof of a house the angle of depression of a stone lying in the field is 60°. If the slant distance of the stone from the roof be 4\(\sqrt {3}\) m then find the height of the house.
Solution:
Here AC = 4\(\sqrt {3}\) m

= 2 × 3 = 6 m
Hence the height of the house = 6 m

Question 3.
The angle of elevation of the top of a pillar from a point on the ground is 15°. On walking 100 m towards the pillar the angle of elevation becomes 30°. Find the height of the pillar.
Solution:
Let the angle of elevation of the top of pillar from pint O on the ground be 15° and from point A be 30°, where OA = 100 m.

∵ ∠OCA = 15° and ∠CAB = 30°
∴ ∠OCA = 15° and OA = AC = 100 m
In right △CBA,
sin 30° = \(\frac{BC}{AC}\)
∴ BC = AC sin 30°
= 100 × \(\frac{1}{2}\) = 50 m
Hence height of the pillar = 50 m

Question 4.
The angle of elevation of the upper end of a pillar from a piont 40 m distant from the base of the pillar is 60°. Find the height of the pillar.
Solution:
Let the height of the pillar be h m

Or h = 40 × \(\sqrt {3}\)
= 40 × 1.732 = 69.280
∴ h = 69.28 m
Hence height of the pillar = 69.28 m

Question 5.
From the points lying at distance x and y in the same line passing through the base of a tower, the angles of elevation of the top of a tower are complementary to each other. Prove that the height of the tower is \(\sqrt {xy}\).
Solution:
Let the height of the tower be h. OB = x and AB = y

Now in right triangle CBO,
tan θ = \(\frac{BC}{OB}\) = \(\frac{h}{x}\)
∴ h = x tan θ ……(i)
Again in right triangle CBA
tan (90° – θ) = \(\frac{h}{x}\)
∴ h = y cot θ
∵ tan (90° – θ) = cot θ …. (ii)
Multiplying equation (i) and equation (ii)
h² = x tan θ × y cot θ = xy
[∵ tan θ × cot θ = 1]
∴ h = \(\sqrt {xy}\)
Hence the height of the tower is = \(\sqrt {xy}\). ( Hence Proved )

Question 6.
A ladder stands against a vertical wall such that its lower end is 3 m distant on the horizontal ground. If the ladder is inclined at an angle of 60° with the ground then find upto what height does the ladder reach the wall? Find also the length of the ladder.
Solution:
Let the ladder AC reach upto height h on the wall AB. Now from right triangle ABC
tan θ = \(\frac{Perpendicular}{base}\) = \(\frac{AB}{BC}\)

∴ h = 3 × \(\sqrt {3}\)
= 3 × 1.732
= 5.196 m
∴ h = 5.196 m
Hence the ladder reaches upto a height of h = 5.196 m on the wall.
For the Length of the ladder—Let the length of the ladder be x m. Therefore from right triangle ABC
cos θ = \(\frac{base}{hypotenuse}\) = \(\frac{BC}{AC}\)
Or cos 60° = \(\frac{3}{x}\)
Or \(\frac{1}{2}\) = \(\frac{3}{x}\)
∴ x = 6 m
Hence length of the ladder = 6 m

Question 7.
An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
Solution:
Here AB is the chimney, CD the observer and ∠ADE the angle of elevation (see figure). Here ADE is a triangle, right-angled at E and we are required to find the height of the chimney.

Here AB = AE + BE
= (AE + 1.5) m
∵ BE = CD = 1.5 m
and DE = CB = 28.5 m
To determine AE, we choose a trigonometric fatio, which involves both AE and DE. Let us choose the tangent of the
angle of elevation.
∴ tan 45° = \(\frac{AE}{DE}\)
Or 1 = \(\frac{AE}{28.5}\)
Therefore AE = 28.5
Hence height of chimney (AB)
= (28.5 + 1.5) m
= 30 m

Question 8.
A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Solution:
In figure AB represents the tower, CB is the distance of the point from the tower and ∠ACB is the angle of elevation. We need to determine the height of the tower, i.e., AB. Also, ACB is a triangle, right-angled at B.
Now we choose the trigonometric ratio tan 60° (Or cot 60°), as the ratio involves AB and BC.

i.e., AB = 15\(\sqrt {3}\)
Hence, the height of the tower is = 15\(\sqrt {3}\) m

Question 9.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 4 m from the banks, find the width of the river.
Solution:
In figure, A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 4 m, i.e.. DP = 4 m. We are interested to detennine the width of the river, which is the length of the side AB of the △APB.

Now AB= AD + DB
In right △APD. ∠A = 30°
So, tan 30° = \(\frac{PD}{AD}\)
i.e., \(\frac{1}{\sqrt{3}}\) = \(\frac{4}{AD}\) Or AD = 4\(\sqrt {3}\) m
Therefore, in right △PBD,
∠B = 45°
So, BD = PD = 4 m
Now AB = BD + AD = 4 + 4\(\sqrt {3}\)
= 4 (1 + \(\sqrt {3}\)) m
Therefore, the width of the river is = 4 (\(\sqrt {3}\) + 1) m

Question 10.
From the top of a building 10 m high, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Solution:
Let the height of the tower be h m. The angle of elevation of the top of the tower is 60° and the angle of depression of the foot is 45°. Now in △ABD

Long Answer Type Questions—

Question 1.
An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder?
Solution:
The electrician is required to reach the point B on the pole AD.

So, BD = AD – AB
= 5 – 1.3 = 3.7 m
Here BC represents the length of the ladder. We need to find the length, i.e., the hypotenuse of the right triangle BDC.
sin 60° = \(\frac{Perpendicular}{hypotenuse}\) = \(\frac{BD}{BC}\)

Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

Question 2.
From the top of a light house 100 m high an observer sees a ship in the sea coming towards him. If the angle of depression of the ship changes to 45° from 30° then find the distance travelled by the ship in this duration of the observer.
Solution:
Let AB be a light house where height is 100- m. Let the ship come to the point C from, D.

Then ∠XAD = 30° = ∠ADB
and ∠XAC = 45° = ∠ACB
In right triangle ABC

(CD + 100) = \(\sqrt {3}\)( 100)
CD = 100\(\sqrt {3}\) – 100
= 100 (\(\sqrt {3}\) – 1)
= 100(1.732 – 1)
CD = 100 × 0.732 = 73.2 m
Hence the distance travelled by , the ship = 73.2 m

Question 3.
A tower stands vertically on a horizontal level and a flag-staff of length 7m stands on the top. From a point on the horizontal level the angles of elevation of the base and the upper end of the flag-staff are 30° and 45° respectively. Find the height of the tower.
Solution:
Let the height of the tower

BD = h m
and the length of the flag-staff
AD = 7 m
From right △ABC

h = \(\frac{7000}{732}\) = 9.56 m
∴ height of the tower = 9.56 m

Question 4.
At the instant when exactly below an aeroplane flying at a height of 4000 m, another also plane comes, the angles of elevation of these aeroplanes from a point on the horizontal level are respectively 60° and 45°. At that instant find the vertical distance between the two aeroplanes.
Solution:
Let the height of the first aeroplane be AC and the height of the second aeroplane be BC = h m.

⇒ h = x . 1
⇒ h = x …..(ii)
From equation (i) and (ii)
h\(\sqrt {3}\) = 4000 m

= \(\frac{4000 \times 1.27}{3}\) = 4000 × 0.4233
= 1693.33 m

Question 5.
The shadow of a tower standing on a level ground is found to be 45 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Solution:
Let the height of the tower AB = h m. Now BD is the shadow of the tower when the angle of elevation of the sun is 30° and BC is the shadow of the tower when the angle of elevation of the Sun is 60°.

Here, DC = 45 m and
BC = x m
∴ BD = 45 + x m
Now from right triangle △ABD

by equality or comparison of equation (i) and (ii)
h\(\sqrt {3}\) – 45 = \(\frac{h}{\sqrt{3}}\)
Or 3h – 45\(\sqrt {3}\) = h
Or 2h = 45\(\sqrt {3}\)
Or h= 22.5 × 1.732
i.e., = 38.97 m
Hence, height of the tower = 38.97 m

Question 6.
From the top and base of a tower 100 m high, the angles of elevation of the top of a cliff are respectively 30° and 45°. Find the height of the cliff.
Solution:
Let the height of the cliff be AB and let it be (h + 100) m and the height of the tower is CD and it is 100 m.

∴ ∠ADE = 30°
and ∠ACB = 45°
From right triangle AED
\(\frac{h}{\mathrm{DE}}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
∵ DE = \(\sqrt {3}\)h …..(i)
and from right triangle ABC
\(\frac{h+100}{\mathrm{BC}}\) = tan 45°
Or \(\frac{h+100}{\mathrm{BC}}\) = 1 [∵ DE = BC]
Or h + 100 = DE
Or h + 100 = \(\sqrt {3}\)h
∵ DE = \(\sqrt {3}\) h from equation (i)
Or h(\(\sqrt {3}\) – 1) = 100
Or h = \(\frac{100}{\sqrt{3}-1}\) × \(\frac{\sqrt{3}+1}{\sqrt{3}+1}\)
[multiplying the numerator and the denominator by (\(\sqrt {3}\) + 1)]
= \(\frac{100(\sqrt{3}+1)}{3-1}\) = \(\frac{100(\sqrt{3}+1)}{2}\)
= 50(1.732 + 1)
= 50 × 2.732
⇒ h = 136.60 m
So, height of the cliff = h + 100
= 136.60 + 100
= 236.60 m

Question 7.
A man on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60°. Find the time taken by the boat to reach the shore.
Solution:
Let the height of the cliff be h m, i.e., according to the figure it is AB and the observer is standing at point A. A boat reaches the point C from point D in 6 minutes.

Let
DC = x₁ CB = x₂
Now from right triangle ABC
\(\frac{h}{x_{2}}\) = tan 60°
⇒ h = x₂ . \(\sqrt {3}\) …….(i)
Now from right triangle ABD
\(\frac{h}{x_{1}+x_{2}}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
⇒ h = \(\frac{x_{1}+x_{2}}{\sqrt{3}}\) ……..(ii)
From equation (i) and (ii)
x₂\(\sqrt {3}\) = \(\frac{x_{1}+x_{2}}{\sqrt{3}}\)
⇒ x₂ × \(\sqrt {3}\) × \(\sqrt {3}\) = x₁ + x₂
⇒ 3x₂ = x₁ + x₂
Or 2x₂ = x₁
Or x₂ = \(\frac{x_{1}}{2}\)
But it is given that in covering the distance x₁, 6 minutes are taken.
∴ Time taken in covering the distance x₂ = \(\frac{1}{2}\) × 6 = 3 minutes
⇒ Time taken in reaching B from point D = 6 + 3 = 9 minutes.

Question 8.
From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take \(\sqrt {3}\) = 1.732)
Solution:
In figure, AB denotes the height of the building, BD the flagstaff and P the given point. There are the right triangles PAB and PAD here. We are required to find the length of the flagstaff i.e., DB and the distance of the building from the point P, i.e., PA.

i.e., the distance of the building from P = 10\(\sqrt {3}\) m = 17.32 m
Let DB = x m
Then,
AD = (10 + x) m
Now in right △PAD

+x = 10\(\sqrt {3}\) – 10 = x
i.e., x = 10(\(\sqrt {3}\) – 1)
= 10 (1.732 – 1)
= 10 × 0.732 = 7.32 m
Hence the length of the flagstaff = 7.32 m.

Question 9.
The shadow of a tower standing on a level ground becomes 40 m longer when the altitude of the Sun reduces to 30° from 60°, i.e., the angle of elevation of the top of the tower from one end of the shadow is 60° and DB is the length of the shadow when the angle of elevation is 30°. Find the height of the tower.

OR
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Solution:
Let the length of AB be h m and BC is x m. According to the question, DB is 40 m longer than BC.
So DB = (40 + x) m
Here, there are two right triangles ABC and ABD.
In △ABC tan 60° = \(\frac{AB}{BC}\)
Or \(\sqrt {3}\) = \(\frac{h}{x}\) …..(i)
In △ABD tan 30 = \(\frac{AB}{BD}\)
i.e., \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+40}\) ……..(ii)
From (i),
h = x\(\sqrt {3}\)
Substituting these values in (ii)
(x\(\sqrt {3}\))\(\sqrt {3}\) = x + 40,
Or 3x = x + 40
Or x = 20
So h = 20\(\sqrt {3}\) [From (i)]
Hence height of the tower = 20\(\sqrt {3}\) m

Question 10.
The angle of depression of the top and the bottom of an 8m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
Solution:
In figure, PC denotes the multi-storeyed building and AB denotes the 8 m tall building. We are interested to determine the height of the nineth-storeyed building, i.e., PC and the distance between the two buildings, i.e.. AC.

PB is a transversal to the parallel lines PQ and BD. Therefore, ∠QPB and ∠PBD arc alternate angles and so are equal. So ∠PBD = 30° Similarly ∠PAC = 45°.
In right △PBD \(\frac{PD}{BD}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
Or BD = PD\(\sqrt {3}\) ….(i)
In right △PAC\(\frac{PC}{AC}\) = tan 45° = 1
i.e., PC = AC
and PC = PD + DC
∴ PD + DC = AC
∵ AC = BD and DC = AB = 8 m,
∴ BD = PD + 8 …. (ii)
From equation (i) and (ii)
PD + 8 = PD\(\sqrt {3}\)
⇒ 8 = PD\(\sqrt {3}\) – PD
⇒ 8 = PD(\(\sqrt {3}\) – 1)

So height of the multi-storey ed building PC
= PD + DC = PD + AB
[∵ DC = AB
= 4(\(\sqrt {3}\) + 1) + 8 | ∵ AB = 8 m
= 4\(\sqrt {3}\) + 4 + 8 = 4\(\sqrt {3}\) + 12
= 4(3 + \(\sqrt {3}\)) m
Distance between the two buildings = AC
But AC = PC
So distance between the two buildings = 4(3 + \(\sqrt {3}\)) m
Hence the height of the multi-storeved building and the distance between the two buildings arc equal to each other.