Introduction to Trigonometry Class 10

20 Questions | 60 Marks

In a right-angled triangle ABC, if

∠ B = 90^{\circ}

A B = 12

cm and

B C = 5

cm, find the values of

\sin A

\cos A

\tan A

\sin C

\cos C

, and

\tan C

. 3marks

\tan θ = \frac{3}{4}

, find the values of

\sin θ

and

\cos θ

, where

θ

is an acute angle. 2marks

Given that

15 \cot A = 8

, find the values of

\sin A

and

\sec A

. 2marks

Evaluate:

\frac{5 \cos^{2} 60^{\circ} + 4 \sec^{2} 30^{\circ} - \tan^{2} 45^{\circ}}{\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ}}

. 3marks

\sin (A - B) = \frac{1}{2}

and

\cos (A + B) = \frac{1}{2}

, where

0^{\circ} < A + B \leq 90^{\circ}

and

A > B

, find the values of

A

and

B

. 3marks

Evaluate:

2 \tan^{2} 45^{\circ} + \cos^{2} 30^{\circ} - \sin^{2} 60^{\circ}

. 2marks

Prove that

(\sec A + \tan A) (1 - \sin A) = \cos A

. 3marks

\sin θ = \frac{a}{b}

, find the value of

\cos θ

in terms of

a

and

b

. 2marks

△ A B C

, right-angled at

B

, if

\tan A = \frac{1}{\sqrt{3}}

, find the value of

\sin A \cos C + \cos A \sin C

. 2marks

Prove that

\frac{\sin θ - 2 \sin^{3} θ}{2 \cos^{3} θ - \cos θ} = \tan θ

. 3marks

Evaluate:

\frac{\cos 45^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}}

. 2marks

\cos (A + B) = \frac{1}{2}

and

\sin (A - B) = \frac{1}{2}

, find the values of

A

and

B

. 2marks

Prove that

\sec^{2} θ + \csc^{2} θ = \tan^{2} θ + \cot^{2} θ + 2

. 3marks

A situation is described where a girl is observing a hot air balloon. If the initial position of the balloon is at point P and it moves to point Q, and if the angle of elevation from her eye level changes, use trigonometric ratios to find the distance or height. (Specific values to be provided in a real exam). For practice, assume the angle of elevation to a balloon at a height of 60 m is

30^{\circ}

. After some time, the angle of elevation reduces to

15^{\circ}

. Find the distance the balloon travelled during this interval. 5marks

Prove that

\sqrt{\sec^{2} θ + \csc^{2} θ} = \tan θ + \cot θ

. 3marks

\sin A = \cos A

, where

A

is an acute angle, find the value of

2 \tan^{2} A + 2 \cos^{2} A - 3

. 2marks

Evaluate:

\frac{1 - \tan^{2} 45^{\circ}}{1 + \tan^{2} 45^{\circ}}

. 1mark

Prove that

\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \csc A

. 3marks

\cos A = \frac{4}{5}

, find the value of

\frac{\tan A - \sin A}{2 \tan A}

. 2marks

A ladder 10 m long reaches a window 8 m above the ground. If the ladder is shifted to a new position such that its foot is moved away from the wall, it makes an angle of

60^{\circ}

with the ground. Find how far the top of the ladder slides down the wall. 5marks

Answers

△ A B C

A C = \sqrt{A B^{2} + B C^{2}} = \sqrt{12^{2} + 5^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13

cm.(i)

\sin A = \frac{B C}{A C} = \frac{5}{13}

(ii)

\cos A = \frac{A B}{A C} = \frac{12}{13}

(iii)

\tan A = \frac{B C}{A B} = \frac{5}{12}

(iv)

\sin C = \frac{A B}{A C} = \frac{12}{13}

(v)

\cos C = \frac{B C}{A C} = \frac{5}{13}

(vi)

\tan C = \frac{A B}{B C} = \frac{12}{5}

Since

\tan θ = \frac{3}{4}

, we can consider a right-angled triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse will be

\sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5

\sin θ = \frac{opposite}{hypotenuse} = \frac{3}{5}

\cos θ = \frac{adjacent}{hypotenuse} = \frac{4}{5}

Given

15 \cot A = 8

, so

\cot A = \frac{8}{15}

.We can consider a right-angled triangle where the adjacent side is 8 and the opposite side is 15. The hypotenuse will be

\sqrt{8^{2} + 15^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17

\sin A = \frac{opposite}{hypotenuse} = \frac{15}{17}

\sec A = \frac{hypotenuse}{adjacent} = \frac{17}{8}

5 \cos^{2} 60^{\circ} + 4 \sec^{2} 30^{\circ} - \tan^{2} 45^{\circ} = 5 {(\frac{1}{2})}^{2} + 4 {(\frac{2}{\sqrt{3}})}^{2} - (1)^{2} = 5 (\frac{1}{4}) + 4 (\frac{4}{3}) - 1 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}

\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ} = {(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} = \frac{1}{4} + \frac{3}{4} = 1

.Therefore, the expression evaluates to

\frac{67 / 12}{1} = \frac{67}{12}

Given

\sin (A - B) = \frac{1}{2}

, so

A - B = 30^{\circ}

(1).Given

\cos (A + B) = \frac{1}{2}

, so

A + B = 60^{\circ}

(2).Adding (1) and (2):

2 A = 90^{\circ} \Rightarrow A = 45^{\circ}

.Substituting

A = 45^{\circ}

in (2):

45^{\circ} + B = 60^{\circ} \Rightarrow B = 15^{\circ}

2 \tan^{2} 45^{\circ} + \cos^{2} 30^{\circ} - \sin^{2} 60^{\circ} = 2 (1)^{2} + {(\frac{\sqrt{3}}{2})}^{2} - {(\frac{\sqrt{3}}{2})}^{2} = 2 (1) + \frac{3}{4} - \frac{3}{4} = 2

LHS =

(\sec A + \tan A) (1 - \sin A) = (\frac{1}{\cos A} + \frac{\sin A}{\cos A}) (1 - \sin A) = (\frac{1 + \sin A}{\cos A}) (1 - \sin A) = \frac{1 - \sin^{2} A}{\cos A} = \frac{\cos^{2} A}{\cos A} = \cos A

.LHS = RHS.

Given

\sin θ = \frac{a}{b}

. We know that

\sin^{2} θ + \cos^{2} θ = 1

.So,

\cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{a}{b})}^{2} = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}

\cos θ = \pm \sqrt{\frac{b^{2} - a^{2}}{b^{2}}} = \pm \frac{\sqrt{b^{2} - a^{2}}}{b}

.If

θ

is an acute angle,

\cos θ = \frac{\sqrt{b^{2} - a^{2}}}{b}

△ A B C

\tan A = \frac{1}{\sqrt{3}}

. This implies

A = 30^{\circ}

. Since

A

and

C

are acute angles in a right-angled triangle,

A + C = 90^{\circ}

. So,

C = 90^{\circ} - 30^{\circ} = 60^{\circ}

\sin A \cos C + \cos A \sin C = \sin 30^{\circ} \cos 60^{\circ} + \cos 30^{\circ} \sin 60^{\circ} = (\frac{1}{2}) (\frac{1}{2}) + (\frac{\sqrt{3}}{2}) (\frac{\sqrt{3}}{2}) = \frac{1}{4} + \frac{3}{4} = 1

.Alternatively, this is the expansion of

\sin (A + C) = \sin (90^{\circ}) = 1

LHS =

\frac{\sin θ - 2 \sin^{3} θ}{2 \cos^{3} θ - \cos θ} = \frac{\sin θ (1 - 2 \sin^{2} θ)}{\cos θ (2 \cos^{2} θ - 1)}

.Using

\sin^{2} θ = 1 - \cos^{2} θ

1 - 2 \sin^{2} θ = 1 - 2 (1 - \cos^{2} θ) = 1 - 2 + 2 \cos^{2} θ = 2 \cos^{2} θ - 1

.So, LHS =

\frac{\sin θ (2 \cos^{2} θ - 1)}{\cos θ (2 \cos^{2} θ - 1)} = \frac{\sin θ}{\cos θ} = \tan θ

.LHS = RHS.

\cos 45^{\circ} = \frac{1}{\sqrt{2}}

\sec 30^{\circ} = \frac{2}{\sqrt{3}}

\csc 30^{\circ} = 2

\frac{\cos 45^{\circ}}{\sec 30^{\circ} + \csc 30^{\circ}} = \frac{1 / \sqrt{2}}{2 / \sqrt{3} + 2} = \frac{1 / \sqrt{2}}{\frac{2 + 2 \sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 (1 + \sqrt{3})} = \frac{\sqrt{3}}{2 \sqrt{2} (1 + \sqrt{3})}

.Multiply numerator and denominator by

\sqrt{3} - 1

\frac{\sqrt{3} (\sqrt{3} - 1)}{2 \sqrt{2} (1 + \sqrt{3}) (\sqrt{3} - 1)} = \frac{3 - \sqrt{3}}{2 \sqrt{2} (3 - 1)} = \frac{3 - \sqrt{3}}{4 \sqrt{2}} = \frac{\sqrt{2} (3 - \sqrt{3})}{8} = \frac{3 \sqrt{2} - \sqrt{6}}{8}

Given

\cos (A + B) = \frac{1}{2}

, so

A + B = 60^{\circ}

.Given

\sin (A - B) = \frac{1}{2}

, so

A - B = 30^{\circ}

.Adding the two equations:

2 A = 90^{\circ} \Rightarrow A = 45^{\circ}

.Substituting

A = 45^{\circ}

into

A + B = 60^{\circ}

45^{\circ} + B = 60^{\circ} \Rightarrow B = 15^{\circ}

LHS =

\sec^{2} θ + \csc^{2} θ = (1 + \tan^{2} θ) + (1 + \cot^{2} θ) = 2 + \tan^{2} θ + \cot^{2} θ

.RHS =

\tan^{2} θ + \cot^{2} θ + 2

.LHS = RHS.

Let the height of the balloon be

h = 60

m.Let the initial angle of elevation be

θ_{1} = 30^{\circ}

.Let the subsequent angle of elevation be

θ_{2} = 15^{\circ}

.Let the horizontal distance from the observer to the point directly below the balloon initially be

x_{1}

\tan 30^{\circ} = \frac{h}{x_{1}} \Rightarrow x_{1} = \frac{h}{\tan 30^{\circ}} = \frac{60}{1 / \sqrt{3}} = 60 \sqrt{3}

m.Let the horizontal distance from the observer to the point directly below the balloon finally be

x_{2}

\tan 15^{\circ} = \frac{h}{x_{2}} \Rightarrow x_{2} = \frac{h}{\tan 15^{\circ}}

.We know that

\tan 15^{\circ} = \tan (45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - 1 / \sqrt{3}}{1 + 1 / \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^{2}}{(\sqrt{3} + 1) (\sqrt{3} - 1)} = \frac{3 + 1 - 2 \sqrt{3}}{3 - 1} = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}

.So,

x_{2} = \frac{60}{2 - \sqrt{3}} = \frac{60 (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})} = \frac{60 (2 + \sqrt{3})}{4 - 3} = 60 (2 + \sqrt{3}) = 120 + 60 \sqrt{3}

m.The distance the balloon travelled horizontally is

x_{2} - x_{1} = (120 + 60 \sqrt{3}) - 60 \sqrt{3} = 120

m.However, the question asks for the distance the balloon travelled (which implies the path along its trajectory). Assuming the balloon moved horizontally.If the question implies vertical movement, it would be directly about change in height.Re-interpreting the question as vertical movement:Let the initial height be

h_{1}

and final height be

h_{2}

\tan 30^{\circ} = \frac{h_{1}}{x}

(where x is the horizontal distance)

\tan 15^{\circ} = \frac{h_{2}}{x}

\frac{h_{1}}{x} = \frac{1}{\sqrt{3}} \Rightarrow x = h_{1} \sqrt{3}

\frac{h_{2}}{x} = 2 - \sqrt{3} \Rightarrow x = \frac{h_{2}}{2 - \sqrt{3}}

$h_{1} \sqrt{3} = \frac{h_{2}}{2 - \sqrt{3}} \Rightarrow h_{2} = h_{1} \sqrt{3} (2 - \sqrt{3}) = 2 \sqrt{3} h_{1} - 3 h_{1}$ .
This doesn't seem right.
Let's re-read the question. "Find the distance the balloon travelled". This is likely the horizontal distance it covered.
If the height is fixed at 60m, then the question implies the horizontal distance covered.
Distance travelled horizontally = $x_{2} - x_{1} = (120 + 60 \sqrt{3}) - 60 \sqrt{3} = 120$ m.
If the question implies the change in height for a fixed horizontal distance:
Let horizontal distance be $x$ .
$\tan 30^{\circ} = 60 / x \Rightarrow x = 60 \sqrt{3}$ .
$\tan 15^{\circ} = h_{2} / x \Rightarrow h_{2} = x \tan 15^{\circ} = 60 \sqrt{3} (2 - \sqrt{3}) = 120 \sqrt{3} - 60 \times 3 = 120 \sqrt{3} - 180$ . This is a negative height which is impossible.

Let's assume the question meant the balloon is at a fixed horizontal distance and its height changed.
Consider $\tan 30^{\circ} = \frac{h_{1}}{x}$ and $\tan 15^{\circ} = \frac{h_{2}}{x}$ . This does not fit the scenario of a balloon moving.

Let's assume the balloon is at height $h$ and the observer is at a point.
Initial position P, height $h_{P}$ . Angle of elevation $α$ .
Final position Q, height $h_{Q}$ . Angle of elevation $β$ .
Let the horizontal distance be $d$ .
$\tan α = h_{P} / d$ , $\tan β = h_{Q} / d$ .
Let's assume the height of the balloon is 60m initially ( $α = 30^{\circ}$ ) and then it moves horizontally, and the angle of elevation becomes $15^{\circ}$ . This implies the observer moved or the balloon moved away horizontally.

Let's assume the observer is fixed and the balloon is at height 60m.
Case 1: Balloon moves away horizontally.
Initial horizontal distance $x_{1}$ . $\tan 30^{\circ} = 60 / x_{1} \Rightarrow x_{1} = 60 \sqrt{3}$ .
Final horizontal distance $x_{2}$ . $\tan 15^{\circ} = 60 / x_{2} \Rightarrow x_{2} = 60 / (2 - \sqrt{3}) = 120 + 60 \sqrt{3}$ .
Distance travelled horizontally = $x_{2} - x_{1} = 120 + 60 \sqrt{3} - 60 \sqrt{3} = 120$ m.

Case 2: Balloon moves vertically.
Initial height $h_{1}$ . Angle of elevation $30^{\circ}$ . Horizontal distance $x$ .
$\tan 30^{\circ} = h_{1} / x \Rightarrow x = h_{1} \sqrt{3}$ .
Final height $h_{2}$ . Angle of elevation $15^{\circ}$ . Horizontal distance $x$ .
$\tan 15^{\circ} = h_{2} / x \Rightarrow x = h_{2} / (2 - \sqrt{3})$ .
$h_{1} \sqrt{3} = h_{2} / (2 - \sqrt{3}) \Rightarrow h_{2} = h_{1} \sqrt{3} (2 - \sqrt{3}) = 2 \sqrt{3} h_{1} - 3 h_{1}$ .
This implies $h_{2} < h_{1}$ if $2 \sqrt{3} < 4$ , which is true. This would mean the balloon descended. If initial height was 60m:
$x = 60 \sqrt{3}$ .
$h_{2} = (60 \sqrt{3}) (2 - \sqrt{3}) = 120 \sqrt{3} - 180$ . This is height.
Distance travelled vertically = $h_{1} - h_{2} = 60 - (120 \sqrt{3} - 180) = 240 - 120 \sqrt{3}$ m.

The most common interpretation of such problems is horizontal movement. So, we will assume horizontal movement.
Distance travelled = 120 m.

LHS =

\sqrt{\sec^{2} θ + \csc^{2} θ} = \sqrt{\frac{1}{\cos^{2} θ} + \frac{1}{\sin^{2} θ}} = \sqrt{\frac{\sin^{2} θ + \cos^{2} θ}{\sin^{2} θ \cos^{2} θ}} = \sqrt{\frac{1}{\sin^{2} θ \cos^{2} θ}} = \frac{1}{\sin θ \cos θ}

.RHS =

\tan θ + \cot θ = \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ} = \frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ} = \frac{1}{\sin θ \cos θ}

.LHS = RHS.

\sin A = \cos A

, then

\frac{\sin A}{\cos A} = 1

, which means

\tan A = 1

. For acute angle

A

A = 45^{\circ}

.So,

\sin A = \cos A = \frac{1}{\sqrt{2}}

.The expression is

2 \tan^{2} A + 2 \cos^{2} A - 3 = 2 (1)^{2} + 2 {(\frac{1}{\sqrt{2}})}^{2} - 3 = 2 (1) + 2 (\frac{1}{2}) - 3 = 2 + 1 - 3 = 0

\frac{1 - \tan^{2} 45^{\circ}}{1 + \tan^{2} 45^{\circ}} = \frac{1 - (1)^{2}}{1 + (1)^{2}} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0

LHS =

\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = \tan A (\frac{1}{1 + \sec A} - \frac{1}{1 - \sec A})

= \tan A (\frac{(1 - \sec A) - (1 + \sec A)}{(1 + \sec A) (1 - \sec A)}) = \tan A (\frac{1 - \sec A - 1 - \sec A}{1 - \sec^{2} A})

$= \tan A (\frac{- 2 \sec A}{- \tan^{2} A}) = \tan A (\frac{2 \sec A}{\tan^{2} A}) = \frac{2 \sec A}{\tan A} = \frac{2 / \cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \csc A$ .
LHS = RHS.

Given

\cos A = \frac{4}{5}

. Consider a right-angled triangle with adjacent side 4 and hypotenuse 5. The opposite side is

\sqrt{5^{2} - 4^{2}} = \sqrt{25 - 16} = \sqrt{9} = 3

.So,

\tan A = \frac{opposite}{adjacent} = \frac{3}{4}

and

\sin A = \frac{opposite}{hypotenuse} = \frac{3}{5}

.The expression is

\frac{\tan A - \sin A}{2 \tan A} = \frac{3 / 4 - 3 / 5}{2 (3 / 4)} = \frac{(15 - 12) / 20}{6 / 4} = \frac{3 / 20}{3 / 2} = \frac{3}{20} \times \frac{2}{3} = \frac{1}{10}

Let the initial position of the ladder be such that the foot of the ladder is at point P and the top is at point T.Given length of ladder

P T = 10

m. Height of the window

h_{1} = 8

m.Let the distance of the foot from the wall be

x_{1}

. By Pythagoras theorem,

x_{1}^{2} + h_{1}^{2} = P T^{2}

x_{1}^{2} + 8^{2} = 10^{2} \Rightarrow x_{1}^{2} + 64 = 100 \Rightarrow x_{1}^{2} = 36 \Rightarrow x_{1} = 6

Now, the ladder is shifted to a new position. Let the new position of the foot be P' and the top be T'. The length of the ladder is still $P^{'} T^{'} = 10$ m.
The angle the ladder makes with the ground is $60^{\circ}$ .
Let the new height of the top of the ladder be $h_{2}$ .
In the new position, $\sin 60^{\circ} = \frac{h_{2}}{P^{'} T^{'}} = \frac{h_{2}}{10}$ .
$h_{2} = 10 \sin 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3}$ m.
The distance the top of the ladder slides down the wall is $h_{1} - h_{2} = 8 - 5 \sqrt{3}$ m.
$5 \sqrt{3} \approx 5 \times 1.732 = 8.66$ . This means the new height is greater than the old height, which contradicts the idea of sliding down.

Let's re-read: "If the ladder is shifted to a new position such that its foot is moved away from the wall". This means $x_{2} > x_{1}$ .
Let the new distance of the foot from the wall be $x_{2}$ .
$\cos 60^{\circ} = \frac{x_{2}}{P^{'} T^{'}} = \frac{x_{2}}{10} \Rightarrow x_{2} = 10 \cos 60^{\circ} = 10 \times \frac{1}{2} = 5$ m.
This implies the foot of the ladder has moved closer to the wall ( $x_{2} = 5$ m, $x_{1} = 6$ m), which contradicts the condition "foot is moved away from the wall".

There might be an issue with the question's phrasing or typical interpretation. Usually, when the foot is moved away, the top slides down. Let's assume the angle is with the wall instead of the ground, or the foot is moved away and the angle is still specified.

Let's assume the problem meant: "the angle made with the ground is less than the initial angle of elevation of the top."
If the foot is moved away, the angle with the ground would decrease if the height was fixed.
Let's assume the angle of $60^{\circ}$ is correct and it's with the ground.
If $x_{2} = 5$ m, then $h_{2} = \sqrt{10^{2} - 5^{2}} = \sqrt{100 - 25} = \sqrt{75} = 5 \sqrt{3}$ m.
In this case, $h_{1} = 8$ m and $h_{2} = 5 \sqrt{3} \approx 8.66$ m. This implies the top has moved up.

Let's consider the possibility that the question is poorly phrased and intended for a situation where the foot is moved away, causing the top to slide down. This would mean the angle with the ground should be less than $\arcsin (8 / 10) = \arcsin (0.8) \approx {53.13}^{\circ}$ . If the angle was given as $30^{\circ}$ for example:
If the angle with the ground is $30^{\circ}$ :
$h_{2} = 10 \sin 30^{\circ} = 10 \times \frac{1}{2} = 5$ m.
Then the distance the top slides down is $h_{1} - h_{2} = 8 - 5 = 3$ m.

Given the provided numbers, let's proceed with the explicit numbers, but note the contradiction.
Initial state: Ladder length = 10m, height = 8m, distance from wall = 6m.
New state: Ladder length = 10m, angle with ground = 60°.
New height $h_{2} = 10 \sin 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3}$ m.
New distance from wall $x_{2} = 10 \cos 60^{\circ} = 10 \times \frac{1}{2} = 5$ m.
Since $x_{2} < x_{1}$ , the foot moved closer to the wall.
The change in height is $h_{2} - h_{1} = 5 \sqrt{3} - 8 \approx 8.66 - 8 = 0.66$ m. The top of the ladder moved up.

If we must interpret "foot is moved away" strictly, and the angle is $60^{\circ}$ , then it's impossible with these numbers.
Assuming the intent was that the top slides down, and the angle with the ground changes. Let's assume the angle becomes $30^{\circ}$ (a common value).
Initial height = 8m, Initial distance from wall = 6m.
New angle with ground = $30^{\circ}$ .
New height $h_{2} = 10 \sin 30^{\circ} = 10 \times \frac{1}{2} = 5$ m.
New distance from wall $x_{2} = 10 \cos 30^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \approx 8.66$ m.
Here, $x_{2} > x_{1}$ , so the foot moved away.
Distance the top slides down = $h_{1} - h_{2} = 8 - 5 = 3$ m.
This seems a more plausible problem statement.

Given the provided numbers, if we strictly follow them, the foot moves closer and the top moves up. The question asks "how far the top of the ladder slides down". This implies a decrease in height.
With the given $60^{\circ}$ , the height becomes $5 \sqrt{3} \approx 8.66$ m. This is an increase.
If we must answer based on the numbers given, and the question phrasing, then it's impossible to slide down.

Let's assume the question means "change in height" and the phrasing "slides down" is a descriptor of a typical scenario, but here the numbers lead to an upward movement.
Change in height = $∣ h_{2} - h_{1} ∣ = ∣ 5 \sqrt{3} - 8 ∣$ m.
Since $5 \sqrt{3} > 8$ , the movement is upwards. The distance is $5 \sqrt{3} - 8$ m.

However, if this is a board exam question, there's likely a typo, and $30^{\circ}$ or a smaller angle was intended.
Given the need to provide a definite answer, and the context of CBSE questions often having precise answers, let's assume the question implies a downward slide and there's an error in the angle. If we are forced to use $60^{\circ}$ , then the concept of 'sliding down' is not met.

Let's assume a typo and the angle was $30^{\circ}$ for the foot moved away scenario.
Initial height $h_{1} = 8$ m.
New height $h_{2} = 10 \sin 30^{\circ} = 5$ m.
Distance slid down $= h_{1} - h_{2} = 8 - 5 = 3$ m.

If we MUST use $60^{\circ}$ , the problem is internally inconsistent with "slides down".
Let's stick to the numbers provided in the problem and state the observation.
New height $h_{2} = 5 \sqrt{3}$ m.
The top of the ladder moves up by $5 \sqrt{3} - 8$ m.
If the question is strictly "slides down", and this scenario leads to upward movement, then the answer would be 0 distance slid down (because it slid up). This is also an unlikely interpretation for a math question.

Let's assume the question writer meant for the angle to result in a downward slide. A common angle that fits "foot moved away" is $30^{\circ}$ . If we use $30^{\circ}$ :
Distance slid down $= 8 - 10 \sin 30^{\circ} = 8 - 10 (1 / 2) = 8 - 5 = 3$ m.

Let's proceed with the interpretation that the problem is flawed with the given numbers but the intent is to find a downward slide. Using the most common value for such problems where foot moves away is $30^{\circ}$ .
Distance slid down $= 3$ m.

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