Class 10 Maths Chapter 9 Exercise 9.1 NCERT Solutions

Question 1 NCERT Ex 9.1

A circus artist is climbing a $20 ext{ m}$ long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^circ$.

Step 1: Geometric Setup

Let vertical pole be $AB$ and rope be hypotenuse $AC = 20 ext{ m}$.
Angle with ground level $angle ACB = 30^circ$. $Delta ABC$ is right-angled at $B$.

Step 2: Apply Trigonometric Ratio ($sin heta$)

In right $Delta ABC$: $$sin 30^circ = rac{ ext{Opposite}}{ ext{Hypotenuse}} = rac{AB}{AC}$$ $$rac{1}{2} = rac{AB}{20} implies AB = rac{20}{2} = 10 ext{ m}$$

The height of the pole is $10 ext{ m}$.

Question 2 NCERT Ex 9.1

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of $30^circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 ext{ m}$. Find the height of the tree.

Step 1: Understand the Broken Tree Components

Let original tree be standing at point $B$, broken at height $C$. The top $A$ bends down to touch ground at $D$.
Standing part $= BC$. Broken part $= CD$ (where $CD = AC$).
Given distance $BD = 8 ext{ m}$, angle $angle BDC = 30^circ$.
Total tree height $= BC + CD$.

Step 2: Compute Standing Part ($BC$)

$$ an 30^circ = rac{BC}{BD} implies rac{1}{sqrt{3}} = rac{BC}{8} implies BC = rac{8}{sqrt{3}} ext{ m}$$

Step 3: Compute Broken Part ($CD$)

$$cos 30^circ = rac{BD}{CD} implies rac{sqrt{3}}{2} = rac{8}{CD} implies CD = rac{16}{sqrt{3}} ext{ m}$$

Step 4: Total Height of Tree

$$ ext{Total Height} = BC + CD = rac{8}{sqrt{3}} + rac{16}{sqrt{3}} = rac{24}{sqrt{3}} = rac{24sqrt{3}}{3} = 8sqrt{3} ext{ m}$$

Total height of the tree is $8sqrt{3} ext{ m}$ (approx $13.86 ext{ m}$).

Question 4 NCERT Ex 9.1

The angle of elevation of the top of a tower from a point on the ground, which is $30 ext{ m}$ away from the foot of the tower, is $30^circ$. Find the height of the tower.

Step 1: Geometric Setup

Let tower height be $h = AB$, distance from foot $BC = 30 ext{ m}$, angle of elevation $angle ACB = 30^circ$.

Step 2: Apply $ an heta$

$$ an 30^circ = rac{AB}{BC} = rac{h}{30}$$ $$rac{1}{sqrt{3}} = rac{h}{30} implies h = rac{30}{sqrt{3}} = rac{30sqrt{3}}{3} = 10sqrt{3} ext{ m}$$

The height of the tower is $10sqrt{3} ext{ m}$ (approx $17.32 ext{ m}$).

Question 5 NCERT Ex 9.1

A kite is flying at a height of $60 ext{ m}$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^circ$. Find the length of the string, assuming that there is no slack in the string.

Step 1: Geometric Setup

Vertical height of kite $AB = 60 ext{ m}$.
Let string length be $AC = L$. Angle of inclination $angle ACB = 60^circ$.

Step 2: Apply $sin heta$

$$sin 60^circ = rac{AB}{AC} = rac{60}{L}$$ $$rac{sqrt{3}}{2} = rac{60}{L} implies L = rac{120}{sqrt{3}} = rac{120sqrt{3}}{3} = 40sqrt{3} ext{ m}$$

The length of the string is $40sqrt{3} ext{ m}$ (approx $69.28 ext{ m}$).

Question 6 NCERT Ex 9.1

A $1.5 ext{ m}$ tall boy is standing at some distance from a $30 ext{ m}$ tall building. The angle of elevation from his eyes to the top of the building increases from $30^circ$ to $60^circ$ as he walks towards the building. Find the distance he walked towards the building.

Step 1: Calculate Effective Height Above Eye Level

Total building height $= 30 ext{ m}$.
Boy's height $= 1.5 ext{ m}$.
Effective vertical height above horizontal eye line $= 30 - 1.5 = 28.5 ext{ m} = rac{57}{2} ext{ m}$.

Step 2: Compute Initial and Final Horizontal Distances

Let initial distance from building be $x$ and final distance be $y$.
1. From initial position ($ heta = 30^circ$): $$ an 30^circ = rac{28.5}{x} implies rac{1}{sqrt{3}} = rac{28.5}{x} implies x = 28.5sqrt{3} ext{ m}$$ 2. From final position ($ heta = 60^circ$): $$ an 60^circ = rac{28.5}{y} implies sqrt{3} = rac{28.5}{y} implies y = rac{28.5}{sqrt{3}} = 9.5sqrt{3} ext{ m}$$

Step 3: Compute Distance Walked

$$ ext{Distance Walked} = x - y = 28.5sqrt{3} - 9.5sqrt{3} = 19sqrt{3} ext{ m}$$

The distance walked towards the building is $19sqrt{3} ext{ m}$ (approx $32.91 ext{ m}$).

Question 9 NCERT Ex 9.1

The angle of elevation of the top of a building from the foot of a tower is $30^circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^circ$. If the tower is $50 ext{ m}$ high, find the height of the building.

Step 1: Geometric Setup

Let tower height $CD = 50 ext{ m}$ and building height $AB = h$.
Let distance between their bases $BD = x$.
From base of building $B$, elevation of tower top $C$ is $60^circ$.
From base of tower $D$, elevation of building top $A$ is $30^circ$.

Step 2: Find Distance $x$ from Tower Triangle $Delta CDB$

$$ an 60^circ = rac{CD}{BD} = rac{50}{x}$$ $$sqrt{3} = rac{50}{x} implies x = rac{50}{sqrt{3}} ext{ m}$$

Step 3: Find Building Height $h$ from Triangle $Delta ABD$

$$ an 30^circ = rac{AB}{BD} = rac{h}{x}$$ $$rac{1}{sqrt{3}} = rac{h}{rac{50}{sqrt{3}}} implies h = rac{50}{sqrt{3} imes sqrt{3}} = rac{50}{3} = 16rac{2}{3} ext{ m} approx 16.67 ext{ m}$$

The height of the building is $rac{50}{3} ext{ m} = 16rac{2}{3} ext{ m}$ (approx $16.67 ext{ m}$).

Question 12 NCERT Ex 9.1

From the top of a $7 ext{ m}$ high building, the angle of elevation of the top of a cable tower is $60^circ$ and the angle of depression of its foot is $45^circ$. Determine the height of the tower.

Step 1: Geometric Setup

Let building be $AB = 7 ext{ m}$ and tower be $CD$.
Draw horizontal line $AE$ parallel to ground $BD$, intersecting tower at $E$.
$ED = AB = 7 ext{ m}$. Let $CE = h$. Total tower height $CD = h + 7$.
Angle of depression to tower foot $D$: $angle EAD = 45^circ$.
Angle of elevation to tower top $C$: $angle CAE = 60^circ$.

Step 2: Use Depression Angle to Find Horizontal Distance ($AE$)

In right $Delta AED$: $$ an 45^circ = rac{ED}{AE} = rac{7}{AE} implies 1 = rac{7}{AE} implies AE = 7 ext{ m}$$

Step 3: Use Elevation Angle to Find Height Component $h$

In right $Delta AEC$: $$ an 60^circ = rac{CE}{AE} = rac{h}{7}$$ $$sqrt{3} = rac{h}{7} implies h = 7sqrt{3} ext{ m}$$

Step 4: Total Height of the Tower

$$ ext{Total Height } CD = CE + ED = 7sqrt{3} + 7 = 7(sqrt{3} + 1) ext{ m}$$ $$= 7(1.732 + 1) = 7(2.732) approx 19.12 ext{ m}$$

The height of the cable tower is $7(sqrt{3} + 1) ext{ m}$ (approx $19.12 ext{ m}$).

Question 16 NCERT Ex 9.1

The angles of elevation of the top of a tower from two points at a distance of $4 ext{ m}$ and $9 ext{ m}$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6 ext{ m}$.

Step 1: Geometric Setup with Complementary Angles

Let tower height be $h = AB$.
Let two observation points on the ground be $C$ and $D$, where $BC = 4 ext{ m}$ and $BD = 9 ext{ m}$.
Since angles of elevation are complementary (sum to $90^circ$):
Let $angle ACB = heta$, then $angle ADB = 90^circ - heta$.

Step 2: Formulate Equations for Both Triangles

From $Delta ABC$: $$ an heta = rac{AB}{BC} = rac{h}{4} quad ext{...(i)}$$ From $Delta ABD$: $$ an(90^circ - heta) = rac{AB}{BD} = rac{h}{9}$$ Using identity $ an(90^circ - heta) = cot heta$: $$cot heta = rac{h}{9} quad ext{...(ii)}$$

Step 3: Multiply Equations (i) and (ii)

$$ an heta cdot cot heta = rac{h}{4} cdot rac{h}{9}$$ Since $ an heta cdot cot heta = 1$: $$1 = rac{h^2}{36} implies h^2 = 36 implies h = sqrt{36} = 6 ext{ m}$$ (Height cannot be negative).

Hence proved that the height of the tower is $6 ext{ m}$.

Applications of Trigonometry - Exercise 9.1

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