These AP 10th Class Maths Chapter Wise Important Questions Chapter 12 Applications of Trigonometry will help students prepare well for the exams.

## AP State Syllabus 10th Class Maths 12th Lesson Important Questions and Answers Applications of Trigonometry

Question 1.

A boy observed the top of an electric pole at an angle of elevation of 30°, when the observation point is 10 meters away from the foot of the pole. Draw suitable diagram for the above situation.

Solution:

AB = Height of the Electric Pole

AC = Distance between the foot of the pole to the observer = 10 m

Angle of Elevation = 30°

Question 2.

Draw a diagram to find the height of the kite in the situation given below. “A person is flying a kite at an angle of elevation ‘a’ and the length of thread from his hand to kite is ‘l'”.

Solution:

B is position of kite.

AB = Length of thread = ‘l’

Persons hand is at ‘A’.

∠BAC – α – Angle of elevation.

Question 3.

A flag pole 4 m tall casts a 6 in shadow. At the same time, a nearby building casts a shadow of 24. m. How tall is the building?

Solution:

∴ The height of the building = 16 cm.

Question 4.

A tower is 100√3 m high.Find the angle of elevation of its top when observed from a point 100 m away from the foot of the tower.

Solution:

Height of a tower

(AB) = 100√3 m.

Distance between foot of tower and observer point (BC) = 100m

InΔABCTan θ = \(\frac{100 \sqrt{3}}{100}\)= √3 = tan 60°

∴ θ = 60°

Question 5.

Rehman observed the top of a temple at an angle of elevation of 30°, when the observation point is 24 m. away from the foot of the temple. Find the height of the temple.

Solution:

InΔABC,

Question 6.

A flag pole 7 m tall, casts a 8 m shadow. At the same time, a nearby building casts a shadow of 32 mts. How tall is the building?

Solution:

ΔABC ~ ΔDEF;

\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\) ⇒ \(\frac{7}{\mathrm{DE}}=\frac{8}{32}\)

∴ DE = 28 m

Question 7.

An observer of height 1.8 m is 13.2 in away from a palm tree. The angle of elevation of the top of tree from his eyes is 45°. What is the height of the palm tree?

Solution:

Height of observer = 1.8 m = AB

Distance from palm tree = 13.2 m = AE = BD

Angle of elevation = ∠CBD = 45°

In Δ BCD = tan 45° = \(\frac{\mathrm{CD}}{\mathrm{BD}}\)

⇒ 1 = \(\frac{\mathrm{CD}}{13.2}\) ⇒ CD = 13.2 m

∴ Height of palm tree = CE

= CD + DE

= 13.2 m + 1.8 = 15 m.

Question 8.

Two poles of heights 6 m. and 11m. stand on a plane ground. If the dis¬tance between the feet of the poles is 12 m. find the distance between their tops.

Solution:

Given,

Height of first pole = AB = 6 m.

Height of second pole = CD = 11 m

Distance between feet of poles = AC = 12 m

Distance between the tops of the pole,

i.e., BD

So, BE = AC = 12 m ’

Similarly, AB EC 6m.

Now, DE = DC – EC = 11 – 6 – 5 m

BD^{2} = DE^{2} + BE^{2}

= 5^{2} + 12^{2} = 25 + 144 – 169

∴ BD = 13 m

Question 9.

An observer flying in an aeroplane at an altitude of 900 m observes two ships in front of him, which are in the same direction at an angles of depression of 60° and 30° respectively. Find the distance between the two ships.

Solution:

Question 10.

A wire of length 24 m had been tied with electric pole at an angle of elevation 30° with the ground. As it is covering a long distance, it was cut and tied at angle of elevation 60° with the ground. How much length of the wire was cut?

Solution:

Let AD be length of wire before cut = 24 m

Let AC be length of wire after cut (AC) = x m

Height of the electric pole = AB

Angle of elevation

∠BDA = 30°

∠BCA = 60°

In right Δ^{le} ABD

sin 30° = \(\frac{\mathrm{AB}}{\mathrm{AD}}\)

\(\frac{1}{2}=\frac{\mathrm{AB}}{24}\) ⇒ 2AB = 24 ⇒ AB = 12m

In right Δ^{le} ABC

sin 60° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) ⇒ \(\frac{\sqrt{3}}{2}=\frac{12}{\mathrm{AC}}\)

= √3 AC = 24 ⇒ AC = \(\frac{24}{\sqrt{3}}\)

= 8√3m = 8 × 1.732 = 13.856 m

Length of the wire was cut

= 24 – 13.856 = 10.144 m

Question 11.

A man on the top of vertical tower observes a car moving at a uniform speed coming direcdy towards it. If it takes 12 seconds to change the angle of depression from 30° to 60°, then how long will the car take to reach the tower from that point.

Solution:

Position of observer = ‘D’

Initial position of car = ‘A’

then angle of depression

= ∠DA = 30°

∴ In ΔACD, ∠C = 30°

⇒ tan 30 = \(\frac{\mathrm{CD}}{\mathrm{AC}}=\frac{1}{\sqrt{3}}\)

⇒ CD = \(\frac{\mathrm{AC}}{\sqrt{3}}\) ……………..(1)

and after 12 seconds of time, position of car = ’B’

then angle of depression

= ∠XDB = 60° = ∠DBC

∴ In ΔBCD, ∠B = 60°

⇒ tan 60° = \(\frac{\mathrm{CD}}{\mathrm{BC}}\) = √3

⇒ CD = BC√3 ……………… (2)

∴ (1) = (2)

⇒ CD = \(\frac{\mathrm{AC}}{\sqrt{3}}\) = BC √3 ⇒ AC = 3BC

Now from the figure

AC = AB + BC

⇒ 3BC = AB + BC ( ∵ AC = 3BC)

⇒ AB = 2BC

So the time taken to cover the distance

\(\overline{\mathrm{AB}}\) or the distance 2\(\overline{\mathrm{BC}}\) =12 seconds

∴ Time taken to cover \(\overline{\mathrm{BC}}\) distance

= 6 sec (∵ \(\frac{12}{2}\))

Then the time taken to approach the tower = time taken to cover the distance

\(\overline{\mathrm{AC}}\) = 3\(\overline{\mathrm{BC}}\) = 3(6) = 18 seconds that means it takes 6 more seconds (18 – 12) to reach tower.

Question 12.

Two boys on either side of a temple of 60 m height observe its top at the angles of elevation 60° and 30°. Find the distance between the two boys.

Solution:

Height of the temple BD = 60 metres.

Angle of elevation of one person ∠BAD = 60° .

Angle of elevation of another person ∠BCD = 30°

Let the distance between the first person and the temple, AD = x and distance between the second person and the temple, CD = d

Question 13.

A pole is arranged from a height of 30m from the ground, making 60° angle with earth. Then what is its length?

Solution:

Height from the earth = AB = 30m

Length of pole = BC = ?

∠ACB = 60°

Then in ΔABC AB BC

Question 14.

A long pole is broken in a storm. Top end of the broken pole touched the head of a man at a distance of ‘d\ Then find the angle between the man and the pole that before storm.

Solution:

AB – height of pole

AC – distance = d

CD = man

‘D’ is head of man

After storm

B’ – broken point

‘D’= ‘B’ (∵ they coincides after storm because D’ i.e., head of man which is touched by end of tree = B’)

∴ AngIe between these two points = zero.

Question 15.

A kite is flown from a building with a height (h) m with a long rope. Now the kite and the person having it are observed with angles of elevation α and β respectively by a boy. The distance between boy and building is ‘x’ m. So draw a diagram for this data.

Solution:

AB is building height = h

K is kites position.

‘C’ is position of boy.

AC = Distance between boy and building AB = X

∠DCK = Elevation angle to kite = ∝

∠BCK Elevation angle to building top = β

Question 16.

A person observes tops of two buildings with an angle of elevations 35° and 46° from the mid point in between them. So which building is higher ? Why?

Solution:

Let ‘E’ is mid point between two buildings AB and CD.

In ΔAEB let ∠E = 35°

tan 35 = \(\frac{\mathrm{AB}}{\mathrm{AE}}\)

∴ Height of building (AB)

= (AE) tan 35° ……………… (1)

In ΔECD let ∠E – 46°

∴ tan 46 = \(\frac{\mathrm{CD}}{\mathrm{EC}}\)

∴ Height of building (CD)

= EC tan 46° ……………. (2)

In (1) and (2) AE = EC

( ∵ E is mid point)

Then value of tan 35 is less than tan 46°. So, height of CD (= EC tan 46) is higher than height of AB (= EC tan 35)

(∵ tan 46 > tan 35)

(∵ values of tan increases from 0 to 90)

Question 17.

A 15m long pole forms 53 m long shadow at 8 AM in the morning. Then find the angle made by sun rays with earth.

Solution:

As shown in figure

Height of pole (AB) = 15m

Length of shadow (BC) = 5√3

∠ACB = θ

∴ In ΔABC tan θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{15}{5 \sqrt{3}}\) = √3

∴ tan θ = √3 ⇒ θ = 60°

∴ The sun rays made 60° angle on the earth at that time.

Question 18.

A right circular cylindrical tower, height ‘h’ and radius ‘r’ stands on the ground. Let ‘P’ be a point in the horizontal plane ground and ABC be the semi-circular edge of the top of the tower such that B is the point in it nearest to P. The angles of elevation of the points A and B are 45° and 60°

respectively. Show that \(\frac{\mathbf{h}}{\mathbf{r}}\) = \(\frac{\sqrt{3}(1+\sqrt{5})}{2}\)

Solution:

As shown in the figure

OA = BD = h (height of cylinder)

OD = r (radius of cylinder) .

‘O’ is the centre.

ABC is the edge of semicircle (in the top of cylinder).

B is nearer to the point P. So B should be at the outer edge of diameter. That means just above ‘D’.

∠DPB = 60°, ∠OPA = 45° (given)

In ΔBDP tan P = \(\frac{\mathrm{BD}}{\mathrm{DP}}\) (P = 60°, BD = h)

⇒ √3 = \(\frac{\mathrm{h}}{\mathrm{DP}}\) ⇒ h = DP√3 …………….(1)

In ΔAB tan P = \(\frac{\mathrm{OA}}{\mathrm{OP}}\)

(here P = 45°, OA = h)

⇒ tan 45° = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) = 1

⇒ OA = OP

So OA = h = OP = OD + DP = r + DP

So h = r + DP …………….. (2)

From the equation (1) & (2)

h = DP√3 , h = r + DP

∴ DP√3 = r + DP

So r = DP√3 – DP

r = DP (√3 – 1) ……………. (3)