**Important questions for Class 10 Maths Chapter 9** Some Applications of Trigonometry are provided here for the board exams of 2020 based on the new pattern of CBSE(NCERT). Students who are preparing CBSE-2020 Maths exam are advised to practice these important questions of Some Applications of Trigonometry For Class 10 to score full marks for the questions from this chapter.

Trigonometry has more applications in our daily existence and hence, the chapter is not only important for the board exam but also useful in many other fields.

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## Important Questions & Answers For Class 10 Maths Chapter 9-Some Applications of Trigonometry

**Q.1: ** **The shadow of a tower standing on level ground is found to be 40 m longer when the Sunâ€™s altitude is 30Â° than when it is 60Â°. Find the height of the tower.**

**Solution:**

Let AB be the tower and BC be the length of its shadow when sunâ€™s altitude (angle of elevation from the top of the tower to the tip of the shadow) is 60Â° and DB be the length of the shadow when the angle of elevation is 30Â°.

Let us assume, AB = h m, BC = x m

DB = (40 +x) m

In right triangle ABC,

tan 60Â° = AB/BC

âˆš3 = h/x

h = âˆš3 xâ€¦â€¦â€¦.(i)

In right triangle ABD,

tan 30Â° = AB/BD

1/âˆš3 =h/(x + 40) â€¦â€¦..(ii)

From (i) and (ii),

x(âˆš3 )(âˆš3 ) = x + 40

3x = x + 40

2x = 40

x = 20

Substituting x = 20 in (i),

h = 20âˆš3

Hence, the height of the tower is 20âˆš3 m.

**Q. 2:** **A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30Â° 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.**

**Solution:**

Using given instructions, draw a figure. Let AC be the broken part of the tree. Angel C = 30 degrees.

BC = 8 m

To Find: Height of the tree, which is AB

From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right Î”ABC,

Using Cosine and tangent angles,

cos 30Â° = BC/AC

We know that, cos 30Â° = âˆš3/2

âˆš3/2 = 8/AC

AC = 16/âˆš3 â€¦(1)

Also,

tan 30Â° = AB/BC

1/âˆš3 = AB/8

AB = 8/âˆš3 â€¦.(2)

From (1) and (2),

Total height of the tree = AB + AC = 16/âˆš3 + 8/âˆš3 = 24/âˆš3 = 8âˆš3 m.

**Q. 3: Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60Â° and 30Â°, respectively. Find the height of the poles and the distances of the point from the poles.**

**Solution:**

Let AB and CD be the poles of equal height.

O is the point between them from where the height of elevation taken. BD is the distance between the poles.

As per the above figure, AB = CD,

OB + OD = 80 m

Now,

In right Î”CDO,

tan 30Â° = CD/OD

1/âˆš3 = CD/OD

CD = OD/âˆš3 … (1)

In right Î”ABO,

tan 60Â° = AB/OB

âˆš3 = AB/(80-OD)

AB = âˆš3(80-OD)

AB = CD (Given)

âˆš3(80-OD) = OD/âˆš3 (Using equation (1))

3(80-OD) = OD

240 – 3 OD = OD

4 OD = 240

OD = 60

Substituting the value of OD in equation (1)

CD = OD/âˆš3

CD = 60/âˆš3

CD = 20âˆš3 m

Also,

OB + OD = 80 m

â‡’ OB = (80-60) m = 20 m

Therefore, the height of the poles are 20âˆš3 m and distance from the point of elevation are 20 m and 60 m respectively.

**Q. 4:** **A pole 6 m high casts a shadow 2âˆš 3 m long on the ground, then the Sunâ€™s elevation is**

**(A) 60Â° (B) 45Â° (C) 30Â° (D) 90Â°**

**Solution:**

Let BC be the pole and AB be its shadow.

In triangle ABC,

tan Î¸ = BC/AB

= 6/2âˆš 3

= 3/âˆš 3

= (âˆš 3 Ã— âˆš 3 )/ âˆš 3

= âˆš 3

tan Î¸ = tan 60Â°

Î¸ = 60Â°

Hence, the Sunâ€™s elevation is 60Â°.

**Q. 5: The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is âˆšst.**

**Solution:**

Let BC = s; PC = t

Let the height of the tower be AB = h.

âˆ ABC = Î¸ and âˆ APC = 90Â° â€“ Î¸

(âˆµ the angle of elevation of the top of the tower from two points P and B are complementary)

In triangle ABC,

tan Î¸ = AC/BC = h/s â€¦â€¦â€¦..(i)

In triangle APC,

tan (90Â° â€“ Î¸) = AC/PC = h/t

cot Î¸ = h/t â€¦â€¦â€¦..(ii)

Multiplying (i) and (ii),

tan Î¸ Ã— cot Î¸ = (h/s) Ã— (h/t)

1 = h^{2}/st

h^{2} = st

h = âˆšst

Hence, the height of the tower is âˆšst.

### Practice Questions For Class 10 Maths Chapter 9 Some Applications of Trigonometry

- The angle of elevation of the top of a hill at the foot of a tower is 60Â° and the angle of elevation of top of the tower from the foot of the hill is 30Â°.If the tower is 50 m high, what is the height of the hill?
- A bridge across the river makes an angle of 45Â° with the river bank. If the length of the bridge across the river is 150 m, what is the width of the river?
- There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are the points directly opposite to each other on the two banks, and in a line with the tree. If the angles of elevation of the top of tree from P and Q are 30Â° and 45Â° respectively, find the height of the tree.
- From a point 20m away from the foot of a tower, the angle of elevation of the top of the tower is 30Â°. Find the height of the tower.
- A flagstaff stands at the top of 5m high tower. From a point on the ground, the angle of elevation of the top of the flagstaff is 60Â° and from the same point, the angle of elevation of the top of the tower is 45Â°. Find the height of the flagstaff.
- A tower subtends an angleÂ Î± at a point A in the place of its base and the angle of depression of the foot of the tower at a point b ft. just above A is Î². Prove that the height of the tower is b tanÂ Î± cot Î².
- A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. From a point on the plane, the angle of elevation of the bottom of the flagstaff is 30Â° and that of the top of the flagstaff is 45Â°. Find the height of the tower.