RS Aggarwal Solutions: Heights and Distances- 2 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

In the figure let AB be the pole and BC is its shadow. Join A and C. We get a right-angled triangle ∆ABC. Angle of elevation of the sun is ∠ACB. Given that, AB = BC. We use trigonometric ratio tan taking AB as height and BC as base to find the angle ∠BCA.
In ∆ABC,
tan ∠ACB = AB/BC = 1
or,

since we know that tanθ is 1 whenever θ = 45°.
So, the correct option is (C).

Let AB be the pole and BC be the shadow. Join A and C. We get a right-angled triangle right angled at B. The angle of elevation of the sun is ∠ ACB (= θ, say). We use the trigonometric ratio tan for the triangle taking AB as height and BC as the base. By the problem, AB = √3BC.
In ∆ABC,

or,
θ = 60°
since we know that tanθ is √3 whenever θ = 60°.
So, the correct option is (C).

Let AB be the tower and BC be the shadow. Join A and C. We get a right-angled triangle right-angled at B. The angle of elevation of the sun is ∠ACB(θ, say). Given that, BC = √3AB. We will use trigonometric ratio tan for the triangle taking AB as height and BC as the base.
In ∆ABC,

or,
θ = 30°
since we know that tanθ is 1/√3 whenever θ = 30°.
So, the correct option is (B).

Let AB be the pole and BC be its shadow. Join A and C. Given that, AB = 12m, BC = 4√3m. We get a right-angled triangle ∆ABC, right angled at B. ∠ACB is the angle of elevation of the sun. We will use the trigonometric ratio tan taking AB as the height and BC as the base.
In ∆ABC,

or,

since we know that tanθ is √3 whenever θ = 60°.
So, the correct option is (A).

All we need to find is the angle of elevation of the sun. In the figure, AB is 5m long stick and BC is 2m long shadow of the stick. We join A and C to get a right-angled triangle ∆ABC. Angle of elevation of the sun is ∠ACB.

In ∆ABC,

or,
tan ∠ACB = 2.5

In this figure, PQ is the tree and QS is its shadow. ∠PSQ is the angle of elevation of the sun. So, ∠PSQ = ∠ACB. Join P and S. We get a right-angled triangle ∆PQS. Given, PQ = 12.5m. We use trigonometric ratio tan taking PQ as the height and QS as the base.
In ∆PQS,

or,

or,

So, the correct option is (D).

In the figure, let AC be the ladder and AB is the wall. BC is the distance of the foot of the ladder from wall AB. Join B and C. We get right-angled triangle ∆ABC. Given that, BC = 2m, ∠ACB = 60°. We use trigonometric ratio tan taking AB as the height and BC as the base.
In ∆ABC,

or,

where cos60° = 1/2.
So, the correct option is (D).

In the figure, let AC be the ladder and AB is the wall. Join B and C. We get a right-angled triangle ∆ABC, right angled at B. Given, AC = 15m, ∠BAC = 60°. We use trigonometric ratio cos taking AC as the hypotenuse and AB as the base. In ∆ABC,

or,
AB = 15 x cos 60° = 15/2 m
So, the correct option is (C).

Let AB be the tower and C be the point on the ground, 30m away from B. Given that, BC = 30m, the angle of elevation, ∠ACB = 30°. Join A, C and B,C. We get a right-angled triangle. We use trigonometric ratio tan taking AB as the height and BC as the base. In ∆ABC,

or,

So, the correct option is (B).

In the figure, AB is the tower and car is at the point C. Join A and B to C. So we get a right-angled triangle ABC with right angle at B. Also height of the tower is 150 m. So, AB = 150 m. Again, draw a line AD parallel to BC. The angle of depression of the car from the top of AB is 30°. So, ∠DAC = ∠ACB = 30°. We have to find the distance of the car from the tower, that is BC. For this, we will use the trigonometric ratio tan in ∆ABC.
So from ∆ABC,

or,

So, the correct option is (B).

Let the kite be at point A. Draw a line perpendicular to the ground at say point B. Since we are assuming no slack in the string, we can assume AC to be the string as in the figure. Join B and C to get a right-angled triangle ABC. Given that the kite is flying 30 m above the ground. So AC = 60 m. So, AB = 30 m. Also the length of the string is 60 m. We are to find the angle made by the string AC with the ground that is, ∠ACB.
In ∆ABC,

Since sin30° =1/2
∠ACB = 30°.
So, the correct option is (B).

Let BD be the cliff and AE be the tower. Join D and E. Now, given that the angle of elevation of the top of the tower is same as the angle of depression of the foot of the tower as seen from the top of the cliff. Join A with B and E with B. Also draw a line BC from B onto AE parallel to DE. We get two right-angled triangles ABC and BDE with right angles at C and D respectively. The angle of elevation of the top of the tower AE from the point B is ∠ABC. And the angle of depression of the foot of the tower AE from the point B is ∠CBE which is same as ∠BED. Given, BD = 20m. We are to find the height of the tower, that is BD. Clearly, BD = CE = 20m, BC = DE. In ∆BCE,

or,

In ∆ABC,

or,
AC = 20m
So, height of the tower is AE = AC + CE = 20m + 20m = 40m
The correct option is (B).

In the above figure, let PD be the girl, AB be the lamp post. Given that, the girl stands at a distance of 3 m from the lamp post. So BP = 3 m. Also the height of the girl is 1.5 m. So PD = 1.5 m. The shadow of the girl cast is of the length 4.5 m. So PC = 4.5 m. Then we have to find the height of the lamp post, that is, AB.
Now, BC = BP + PC = 3 + 3.5 = 7.5 m.
From ∆DPC,

∠ACB = ∠DCP. So, 
From, ∆ABC,

or,

or,

Hence the correct choice is C.

In the above figure, let AB be the tower and BC and BD are the consecutive shadows of AB for two different positions of the sun.
From ∆DBA,

⇒ 1 = h/y
⇒ h = y
or,
DB = BA
Now, from ∆ACB,

or, 
⇒  
On cross multiplying we get,
⇒ 2x = √3(h+y)
⇒ 2x + y = √3
⇒ 2x + h = (√3h)
⇒ 2x = (√3 – 1)h
⇒ x = (√3 – 1)h
 
⇒ h = (√3 + 1)x

Let AB be the vertical rod and BC be its shadow. Given that,

We have to find ∠ACB.
We have,

So,
∠ACB = 30°
So the correct choice is A.

In the above figure, let AB be the pole and BC be its shadow. Given that BC = 2√3 m and ∠ACB = 60°. We have to find AB.

or,

or,
AB = 2√3 × √3 = 2 × 3 = 6 m
Hence the correct choice is B.

Given, AB = 20 m, its shadow BC = 20√3 m. We have to find ∠ACB.

So,
∠ACB = 30°
The correct choice is A.

Let BC = x and AD = y be the two towers. They subtend angles 30° and 60° at the centre of the line joining their feet say P. Join C, P, D. Also join B, P and A, P. So, ∠BPC = 30° and ∠APD = 60°. Also, we get two right-angled triangles BPC and APD with right angles at C and D respectively. We have to find x:y.
From ∆BPC,
tan ∠BPC = BC/PC
or,

or,

From ∆APD,

or,

or,
y = PD√3
So, taking the ratio we get,

Hence, the correct choice is C.

In the above figure, let AB be the tower and C be the point on the ground 30 m away from the foot of AB from where the point A is observed. Join B, C and A, C. We get a right-angled triangle ABC with right angle at B. BC = 30 m. Also, the angle of elevation of the top of the tower AB from C is 30°. So, ∠ACB = 30°. We need to find the height of the tower, that is, AB.
In ∆ABC,

or,

or,

Hence the correct choice is B.

In the above figure, let A be the position of the kite. Draw a perpendicular AB from A onto the ground. Since there is no slack in the string, so we can take AC = 100 m to be the string, where C is the position where the string touches the ground. Also, the string makes an angle of 60° with the horizontal. Join B and C. So we get a right-angled triangle ABC with right angle at B and ∠ACB = 60°. We have to find the height of the kite from the ground, that is, AB.

or,

or,

Hence the correct choice is A.

In the above figure, let AB be the tower and C and D be the two points on the ground from where A is observed. Join B, C, D. Let BC = a and BD = b. Join C, D with A. We get two right-angled triangles ABC and ABD with right angle at B. Also, the angles of elevation of the top of the tower AB from C and D are ∠ACB and ∠ADB respectively. These angles are complementary. So,   We are to find the height of the tower, that is, AB.
From ∆ABD,

Again, from ∆ABC,

or,

or,

Now, multiplying we get,

or,
AB² = ab
or,
AB = √ab
So the correct choice is (B).

Let AB be the tower and C, D are two points on the ground. Join B, C, D. Also, join A with C and D. We get two right-angled triangles ABC and ABD with right angle at B. The angles of elevation of the top of the tower from the points D and C are 30° and 60° respectively. So, ∠ADB = 30° and ∠ACB = 30°. We are to find the height of the tower. Also given that the distance from D to C is 20 m. So CD = 20 m.
In ∆ABC,

or,

In ∆ABD,

or,

or,

or,
AB = 10√3m
So, the correct option is (B).

Let ABCD be the rectangle. Given that, BD = 8cm, ∠DBC = 30°.
In ∆BCD,

or,
DC = 8 x sin30º = 4cm.
Again,

or,
BC = 8 x cos 30° = 4√3m
Area of the rectangle = .
So, the correct option is (C).

Let AB be the hill and C, D are the km stones. Join B, C, D. Also join C, D with A. So we get two right-angled triangles ABC and ABD with right angle at B. Draw a line AE parallel to BD. Given that the angles of depression of C and D are 45° and 30° respectively. So, ∠EAD = ∠ADB = 30° and ∠EAC = ∠ACB = 45°. Now, CD = 1km. We are to find the height of the hill, that is, AB.
In ∆ABC,
or,
AB = BC
In ∆ABD,

or,

or,

So, the correct option is (B).

Let AB be the pole and BD, BC is its shadows when the angle of elevation of the sun is 30°, 60° respectively. Here, the height of the pole is 15 m. So, AB = 15m. Join C and D with A. So we get two right-angled triangles ABC and ABD with right angle at B. And ∠ADB = 30° and ∠ACB = 60°. We are to find the difference between the length of the shadows, that is, CD. For this, we find the lengths BD and BC from triangles ABD and ABC respectively using the trigonometric ratio tan. In ∆ABC,

or,

In ∆ABD,

or,
CD = 15√3 -5√3 = 10 √3 m.
So, the correct option is (C).

Let AE be the tower and CD be the observer. Join E and D. The observer is 28.5 m away from the tower. So, ED = 28.5 m. The observer is 1.5 m tall. So, CD = 1.5 m. Given that the angle of elevation of the top of the tower from the eye of the observer is 45°. Join A and C. Also draw a line BC from C onto AE parallel to DE. We get a right-angled triangle ABC with right angle at B and ∠ ACB = 45°. We are to find the height of the tower, that is, AE.
We see that, BC = ED = 28.5 m.
In ∆ABC,

or,
AB = BC = 28.5m
So, the height of the tower is AE = EB + BA = 1.5m + 28.5m = 30m.
The correct option is (B).