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Mathematics · Some Applications of Trigonometry

NCERT Examples

All 7 worked examples from Some Applications of Trigonometry, with independently verified solutions and direct navigation.

Mathematics · Chapter 9 · NCERT Examples

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NCERT Example

Example 1

angle of elevation · tower height

A tower stands vertically on the ground. From a point 15 m away from the foot of the tower, the angle of elevation of its top is 60°. Find the height of the tower.

Tower AB, point C on the ground 15 metres from its foot B, and a 60 degree angle of elevation to A.
Figure 9-4 from the supplied NCERT chapter.

Solution

Board-exam working:

Let AB be the height of the tower and BC = 15 m.

tan⁡60∘=ABBC\tan 60^\circ=\frac{AB}{BC}
3=AB15\sqrt3=\frac{AB}{15}
AB=153 mAB=15\sqrt3\text{ m}

Final answer

The height of the tower is 15√3 m.

NCERT Example

Example 2

ladder · angle of elevation

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. What length of ladder, inclined at 60° to the horizontal, is required? How far from the foot of the pole should the ladder be placed? Take √3 = 1.73.

A ladder BC leaning against pole AD to reach point B, making a 60 degree angle with the ground.
Figure 9-5 from the supplied NCERT chapter.

Solution

Board-exam working:

The height reached on the pole is:

BD=5−1.3=3.7 mBD=5-1.3=3.7\text{ m}

Let BC be the ladder.

sin⁡60∘=BDBC\sin60^\circ=\frac{BD}{BC}
32=3.7BC\frac{\sqrt3}{2}=\frac{3.7}{BC}
BC=3.7×23=7.41.73≈4.28 mBC=\frac{3.7\times2}{\sqrt3}=\frac{7.4}{1.73}\approx4.28\text{ m}

Let DC be the distance of the ladder from the pole.

tan⁡60∘=BDDC\tan60^\circ=\frac{BD}{DC}
3=3.7DC\sqrt3=\frac{3.7}{DC}
DC=3.71.73≈2.14 mDC=\frac{3.7}{1.73}\approx2.14\text{ m}

Final answer

Required ladder length ≈ 4.28 m; distance from the pole ≈ 2.14 m.

NCERT Example

Example 3

observer height · chimney

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°. Find the height of the chimney.

Observer CD viewing the top A of chimney AB at an angle of elevation of 45 degrees.
Figure 9-6 from the supplied NCERT chapter.

Solution

Board-exam working:

Let AE be the part of the chimney above the observer's eye level.

DE=CB=28.5 mDE=CB=28.5\text{ m}
tan⁡45∘=AEDE\tan45^\circ=\frac{AE}{DE}
1=AE28.51=\frac{AE}{28.5}
AE=28.5 mAE=28.5\text{ m}
AB=AE+EB=28.5+1.5=30 mAB=AE+EB=28.5+1.5=30\text{ m}

Final answer

The chimney is 30 m high.

NCERT Example

Example 4

flagstaff · two angles of elevation

From a 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 and the angle of elevation of the top of the flagstaff is 45°. Find the flagstaff length and the distance of the building from P. Take √3 = 1.732.

A 10 metre building AB with flagstaff BD, viewed from point P at angles of 30 and 45 degrees.
Figure 9-7 from the supplied NCERT chapter.

Solution

Board-exam working:

Let PA be the distance from P to the building.

tan⁡30∘=ABPA\tan30^\circ=\frac{AB}{PA}
13=10PA\frac1{\sqrt3}=\frac{10}{PA}
PA=103=17.32 mPA=10\sqrt3=17.32\text{ m}

Let DB = x m be the flagstaff length. Then AD = 10 + x.

tan⁡45∘=ADPA\tan45^\circ=\frac{AD}{PA}
1=10+x1031=\frac{10+x}{10\sqrt3}
10+x=10310+x=10\sqrt3
x=10(3−1)=7.32 mx=10(\sqrt3-1)=7.32\text{ m}

Final answer

The flagstaff is 7.32 m long and the building is 17.32 m from P.

NCERT Example

Example 5

shadows · altitude of sun

The shadow of a tower is 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.

Tower AB casting shadows BC and BD when the Sun's altitudes are 60 and 30 degrees.
Figure 9-8 from the supplied NCERT chapter.

Solution

Board-exam working:

Let AB = h m and BC = x m. Then BD = x + 40 m.

Using triangle ABC:

tan⁡60∘=ABBC\tan60^\circ=\frac{AB}{BC}
3=hx\sqrt3=\frac{h}{x}
h=x3(1)h=x\sqrt3\qquad\text{(1)}

Using triangle ABD:

tan⁡30∘=ABBD\tan30^\circ=\frac{AB}{BD}
13=hx+40\frac1{\sqrt3}=\frac{h}{x+40}
h=x+403(2)h=\frac{x+40}{\sqrt3}\qquad\text{(2)}

Equating (1) and (2):

x3=x+403x\sqrt3=\frac{x+40}{\sqrt3}
3x=x+403x=x+40
x=20x=20
h=203 mh=20\sqrt3\text{ m}

Final answer

The height of the tower is 20√3 m.

NCERT Example

Example 6

angles of depression · two buildings

The angles of depression of the top and bottom of an 8 m 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 buildings.

An 8 metre building AB and multi-storeyed building PC with 30 and 45 degree angles of depression.
Figure 9-9 from the supplied NCERT chapter.

Solution

Board-exam working:

Let PD = x m. Since AB = DC = 8 m, the total height PC = x + 8 m.

In triangle PBD:

tan⁡30∘=PDBD\tan30^\circ=\frac{PD}{BD}
13=xBD\frac1{\sqrt3}=\frac{x}{BD}
BD=x3(1)BD=x\sqrt3\qquad\text{(1)}

In triangle PAC:

tan⁡45∘=PCAC=1\tan45^\circ=\frac{PC}{AC}=1
AC=PC=x+8(2)AC=PC=x+8\qquad\text{(2)}

The distance between the buildings is the same, so AC = BD.

x+8=x3x+8=x\sqrt3
x(3−1)=8x(\sqrt3-1)=8
x=83−1=4(3+1)x=\frac8{\sqrt3-1}=4(\sqrt3+1)
PC=x+8=4(3+3) mPC=x+8=4(\sqrt3+3)\text{ m}
AC=4(3+3) mAC=4(\sqrt3+3)\text{ m}

Final answer

Height of the multi-storeyed building = 4(√3 + 3) m; distance between the buildings = 4(√3 + 3) m.

NCERT Example

Example 7

bridge · angles of depression

From a point on a bridge across a river, the angles of depression of the opposite banks are 30° and 45°. If the bridge is 3 m above the banks, find the width of the river.

Bridge point P three metres above river banks A and B, with angles of depression 30 and 45 degrees.
Figure 9-10 from the supplied NCERT chapter.

Solution

Board-exam working:

Let P be the observation point and PD = 3 m. The river width is AB = AD + DB.

In triangle APD:

tan⁡30∘=PDAD\tan30^\circ=\frac{PD}{AD}
13=3AD\frac1{\sqrt3}=\frac3{AD}
AD=33 mAD=3\sqrt3\text{ m}

In triangle PBD:

tan⁡45∘=PDDB\tan45^\circ=\frac{PD}{DB}
1=3DB1=\frac3{DB}
DB=3 mDB=3\text{ m}
AB=AD+DB=33+3=3(3+1) mAB=AD+DB=3\sqrt3+3=3(\sqrt3+1)\text{ m}

Final answer

The river is 3(√3 + 1) m wide.