NCERT Class 10 Math Chapter 9 Important Questions Answer - Some Application of Trigonometry

Class	10
Chapter	Some Application of Trigonometry
Subject	Math
Category	Important Question Answer

Class 10 Math Chapter 9 Important Question Answer

Q1. The shadow of a tower standing on a level ground is found to be 40 m longer when the altitude (the angle of elevation) of sun changes from 60° to 30°. Find the height of tower.

Ans. Let AB be the tower and BC be the length of the shadow when the sun’s altitude is 60° and DB is the length of the shadow, when the angle of elevation is 30°.

Now, Let AB = h m, and BC = x m.

According to Question,

DB = (40 + x) m

In right ∆ ABC, tan60° = $\displaystyle \frac{{AB}}{{BC}}$

$\displaystyle \sqrt{3}=\frac{h}{x}$

h = $\displaystyle \sqrt{3}x$ —- (i)

In right ∆ ABD, tan30° = $\displaystyle \frac{{AB}}{{BD}}$

$\displaystyle \frac{1}{{\sqrt{3}}}=\frac{h}{{x+40}}$ ——- (ii)

Putting value of eq (i) in eq(ii) we get,

$\displaystyle \frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}x}}{{x+40}}$

3x = x + 40

2x = 40

x = 20

h = $\displaystyle 20\sqrt{3}$

Therefore, the height of the tower is $\displaystyle 20\sqrt{3}$ m.

Q2. The angle of elevation of the top of a building from foot of tower is 30° and angle of elevation of top of the tower from the foot of building is 60°. If height of tower is 50 m. Find the height of building.

Ans.

Let AB be the building and DC be the tower.

Given, Height of tower DC = 50m,

∠CBD = 60° and ∠BCA = 30°

Height of building AB = ?

In right ∆ DCB,

tan60° = $\displaystyle \frac{{DC}}{{BC}}$

BC = $\displaystyle \frac{{50}}{{\sqrt{3}}}$ —- (i)

In right ∆ ABC,

tan30° = $\displaystyle \frac{{AB}}{{BC}}$

Putting value of BC in eq, we get

$\displaystyle \frac{1}{{\sqrt{3}}}=\frac{{AB}}{{\frac{{50}}{{\sqrt{3}}}}}$

3AB = 50

AB = $\displaystyle \frac{{50}}{3}$

Therefore, Height of the tower is $\displaystyle \frac{{50}}{3}$ m.

Q3. From the top of a 7 m building the angle of elevation of the top of a tower is 60° and angle of depression is 45°. Find the height of the tower. Most Important

Ans.

Given, height of building AB = CD = 7m

∠DAC = ∠BAC = 45° (alternate angles)

Also, ∠DAE = 60°

Let height of tower be EC m.

In right ∆ ABC,

tan45° = $\displaystyle \frac{{AB}}{{BC}}$

$\displaystyle 1=\frac{7}{{BC}}$

BC = 7 m

Also, BC = AD

In right ∆ ADE,

tan60° = $\displaystyle \frac{{ED}}{{AD}}$

$\displaystyle \sqrt{3}=\frac{{ED}}{7}$

ED = $\displaystyle 7\sqrt{3}$ m

Height of tower EC = ED + DC = $\displaystyle 7\sqrt{3}$ + 7 = $\displaystyle 7\left( {\sqrt{3}+1} \right)$ m

Therefore, height of tower is $\displaystyle 7\left( {\sqrt{3}+1} \right)$ m.

Q4. The angle of elevation of the top of a tower from a point on the ground, which is 20 m away from the foot of the tower, is 30°. Find the height of the tower.

Ans.

Let height of tower be AB m.

Given, distance of a point from the foot of tower BC = 20 m.

Also, angle of elevation of the top of tower = 60°

In right ∆ ABC,

tan45° = $\displaystyle \frac{{AB}}{{BC}}$

$\displaystyle 1=\frac{{AB}}{{20}}$

AB = 20

Therefore, height of the tower is 20m.

Q5. An observer 1.6m tall is 20m away from a tower. The angle of elevation of the top of the tower from his eyes is 60°. Find the height of the tower? Most Important

Ans.

Given Height of girl AB = CD = 1.6m,

Distance between tower and girl AD = BC = 20 m

Also, angle of elevation of the top of the the tower from girl eyes i.e ∠DAE = 60°

From the fig, Let height of tower be CE m.

In right ∆ ADE,

tan60° = $\displaystyle \frac{{DE}}{{AD}}$

$\displaystyle \sqrt{3}=\frac{{DE}}{{20}}$

DE = $\displaystyle 20\sqrt{3}$

Now, CE = CD + DE = 1.6 + $\displaystyle 20\sqrt{3}$

Therefore, height of the tower is ( $\displaystyle 1.6+20\sqrt{3}$ ) m.

Q6. 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.

Ans.

Let the height of tower AB m.

Given, distance of a point from ground BC = 15 m

Angle of elevation of the top of the tower is 60°.

In right ∆ ABC,

tan60° = $\displaystyle \frac{{AB}}{{BC}}$

$\displaystyle \sqrt{3}=\frac{{AB}}{{15}}$

AB = $\displaystyle 15\sqrt{3}$

Therefore, height of the tower is $\displaystyle 15\sqrt{3}$ m.

Q7. A wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that wire will be taut ?

Ans.

let the pole be BC and length of wire be AB. To keep the wire taut let it be fixed to stake at A.

Then, ABC is right angle triangle at C.

Also, BC = 18m, AB = 24m.

AC is distance between pole base to that point.

In right ∆ ACB,

Using pythagoras theroem

AB² = AC² + BC²

24² = AC² + 18²

576 – 324 = AC²

252 = AC²

AC = $\displaystyle \sqrt{{252}}=6\sqrt{7}$ m

Therefore, required distance is $\displaystyle 6\sqrt{7}$ m.

Q8. A ladder is placed against a wall such that its foot is at a distance of 2.5m from the wall and its top reaches a window 6m above the ground. Find the length of the ladder.

Ans.

Given, height of window AB = 6 m

Distance between foot of ladder and wall BC = 2.5 m

Let the length of ladder be AC m.

In right ∆ ABC,

Using Pythagoras theorem

AB² + BC² = AC²

6² + 2.5² = AC²

36 + 6.25 = AC²

AC² = 42.25

AC = 6.5

Therefore, length of ladder is 6.5 m.

Q9. A vertical pole of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.

Ans.

Height of pole=AB=6 m

Length of shadow of pole =BC=4 m

Length of shadow of tower=EF=28 m

In △ABC and △DEF

∠B=∠E=90° both 90° as both are vertical to ground

∠C=∠F (same elevation in both the cases as both shadows are cast at the same time)

∴△ABC∼△DEF by AA similarity criterion

We know that if two triangles are similar, ratio of their sides are in proportion

So, $\displaystyle \frac{{AB}}{{DE}}=\frac{{BC}}{{EF}}$

$\displaystyle \frac{6}{{DE}}=\frac{4}{{28}}$

DE = 6×7 = 42 m

Hence the height of the tower is 42 m.

Q10. A person goes 10 m due east and then 30 m due north. Find the distance from the starting point.

Ans.

Let the distance from the starting to ending point be AC.

Given AB = 10 m, BC = 30 m.

In right ∆ ABC,

Using Pythagoras theorem

AB² + BC² = AC²

102 + 302 = AC2

1000 = AC2

AC = $\displaystyle \sqrt{{1000}}=10\sqrt{{10}}$ m.

Therefore, required distance is $\displaystyle 10\sqrt{{10}}$ m.

Q11. A ladder 17 m long reaches a window of a building 15 m above the ground, find the distance of the foot of the ladder from the building.

Ans.

Given, length of ladder AC = 17m

Height of building AB = 15 m

Let the distance of the foot of the ladder from the building be BC m.

In right ∆ ABC,

Using Pythagoras theorem

AB² + BC² = AC²

15² + BC² = 17²

289 – 225 = BC²

64 = BC²

BC = 8

Therefore, the distance of foot of the ladder from the building is 8 m.

Q12. Two poles of heights 6 m and 12 m stand on a level plane ground. If the distance between the feet of the poles is 8 m, then find the distance between their tops.

Ans.

Given, Height of first pole AD = 6m

Height of other pole BE = 12m

Distance between two poles AB = DC = 8 m

Let distance between poles tops be DE.

As, AB = DC = 8m

AD = BC = 6m

So, EC = EB – BC = 12 – 6 = 6m

Now, In right ∆ DCE,

Using Pythagoras theorem

DC² + EC² = DE²

8² + 6² = DE²

64 + 36 = DE²

DE = 10

Therefore, distance between tops of poles is 10 m.

Q13. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.

Ans.

In the fig., BC denotes the height of building, CD the tower and A the given point.

Considering right ∆ ABC,

$\displaystyle \tan 45{}^\circ =\frac{{BC}}{{AB}}$

$\displaystyle 1=\frac{{20}}{{AB}}$

AB = 20m

Now, In right ∆ ABD,

$\displaystyle \tan 60{}^\circ =\frac{{BD}}{{AB}}$

$\displaystyle \sqrt{3}=\frac{{BD}}{{20}}$

BD = $\displaystyle 20\sqrt{3}$ m

DC = DB – BC = $\displaystyle 20\sqrt{3}$ – 20 = $\displaystyle 20\left( {\sqrt{3}-1} \right)$ m

Therefore, height of tower is $\displaystyle 20\left( {\sqrt{3}-1} \right)$ m.

Q14. 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 3 m from the banks, find the width of the river.

Ans.

In Fig, 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 poin on the bridge at a height of 3 m, i.e., DP = 3m.

Width of the river AB = ?

Now, AB = AD + DB

In right ∆ APD, ∠A = 30°

So, tan30° = $\displaystyle \frac{{PD}}{{AD}}$

$\displaystyle \frac{1}{{\sqrt{3}}}=\frac{3}{{AD}}$

AD = $\displaystyle 3\sqrt{3}$ m

Also, in right ∆ PBD, ∠B = 45°. So, BD = PD = 3m.

Now, AB = BD + AD = 3 + $\displaystyle 3\sqrt{3}$ = $\displaystyle 3\left( {1+\sqrt{3}} \right)$ m.

Therefore, the width of the river is $\displaystyle 3\left( {1+\sqrt{3}} \right)$ m.

Q15. An electrician has to repair an electric fault on a pole of height 5 m. He 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 he should use which, when inclined at an angle of 60° to the horizontal, would enable him to reach the required position ? (Take = 1.73)

Ans.

In fig., the electrician is required to reach the point B on the pole AD.

BD = AD – AB = 5 – 1.3 = 3.7 m

Here, BC represent the ladder.

In right ∆ BDC,

$\displaystyle \sin 60{}^\circ =\frac{{BD}}{{BC}}$

$\displaystyle \frac{{\sqrt{3}}}{2}=\frac{{3.7}}{{BC}}$

$\displaystyle BC=\frac{{3.7\times 2}}{{\sqrt{3}}}=4.28m$ (approx.)

Therefore, the length of ladder should be 4.28 m.

Q16. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Most Important

Ans.

In Fig., BD is the height of tree and from point C tree break down and touches the ground at point A. So, BD = BC + CD and DC = AC

Distance between foot of tree and point where top touches the ground is AB = 8m

In right ∆ ABC, ∠A = 30°

$\displaystyle \tan 30{}^\circ =\frac{{BC}}{{AB}}$

$\displaystyle \frac{1}{{\sqrt{3}}}=\frac{{BC}}{8}$

BC = $\displaystyle \frac{8}{{\sqrt{3}}}=\frac{{8\sqrt{3}}}{3}$ m

Also, $\displaystyle \sin 30{}^\circ =\frac{{BC}}{{AC}}$

$\displaystyle \frac{1}{2}=\frac{{\frac{{8\sqrt{3}}}{3}}}{{AC}}$

AC = $\displaystyle \frac{{16\sqrt{3}}}{3}$ m = CD

BD = BC + CD = $\displaystyle \frac{{8\sqrt{3}}}{3}+\frac{{16\sqrt{3}}}{3}=\frac{{24\sqrt{3}}}{3}=8\sqrt{3}$ m

Therefore, height of tree (before broken) is $\displaystyle 8\sqrt{3}$ m.

Q17. From the point P on the ground the angle of elevation of the top of a 10 m building is 30°. A flag is hoisted at the top of building and the angle of elevation of the top of flagstaff from P is 45°. Find the length of flagstaff and distance of building from P.

Ans.

In Fig., AB denotes the height of the building, BD the flagstaff and P the given point.

BD = ?

Considering right ∆ PAB,

$\displaystyle \tan 30{}^\circ =\frac{{AB}}{{AP}}$

$\displaystyle \frac{1}{{\sqrt{3}}}=\frac{{10}}{{AP}}$

AP = $\displaystyle 10\sqrt{3}$

The distance of the building from P is $\displaystyle 10\sqrt{3}$ m = 17.32 m.

Now, let us suppose DB = x m. Then AD = (10 + x) m.

Now, in right ∆ PAD, $\displaystyle \tan 45{}^\circ =\frac{{AD}}{{AP}}$

$\displaystyle 1=\frac{{10+x}}{{10\sqrt{3}}}$

x = $\displaystyle 10(\sqrt{3}-1)$ = 7.32

So, the length of the flagstaff is 7.32 m.

Q18. A 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. Most Important

Ans.