Height and Distance
- The angles of depression of two ships from the top of a light house are 45° and 30° towards east. if the ships are 200 m apart, find the height of the light house .
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Let us draw a figure below as per given question.
Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ∠ADC = 30° ∠ACB = 45° and BC = x meter (say)
Correct Option: D
Let us draw a figure below as per given question.
Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ∠ADC = 30° ∠ACB = 45° and BC = x meter (say)
Now from right triangle ABC,
tan 45° = h/x ⇒ 1 = h/x
∴ x = h
Again from right triangle ABD,
tan 30° = h/( 200 + x )
⇒ 1/√3 = h/( 200 + x )
Since x = h , we will get.
⇒ 1/√3 = h/(200 + h )
⇒ (200 + h ) = h X √3
⇒ h X √3 - h = 200
⇒ h x 1.732 - h = 200
⇒ h(1.732 - 1) = 200
∴ h = 200/0.732 = 273.2 m = 273 m
- At a point on a level plane a tower subtends an angle Θ and a flag-staff a ft. in length at the top of the tower subtends an angle Θ . The height of the tower is :
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Let us draw a figure below as per given question.
Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that ∠OAP = Θ and ∠PAQ = Φ
In ΔOAP and ΔOAQ, Apply the formula
TanΘ = P/B = Perpendicular distance / Base distance
Proceed further and solve the given question by applying the above formula.
Correct Option: B
Let us draw a figure as per given question.
Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that ∠OAP = Θ and ∠PAQ = Φ
In ΔOAP use the formula
TanΘ = P/B = Perpendicular distance / Base distance
⇒ TanΘ = OP/OA
⇒ OA = OP CotΘ
Put the value of OP in above equation, we will get
⇒ OA = hCotΘ ................ (1)
Now from triangle ΔOAQ,
⇒ Tan(Θ + Φ) = OQ/OA
⇒ OA = OQ Cot(Θ + Φ)
⇒ OA = (h + a) Cot(Θ + Φ) .............(2)
from equation (1) and (2), We will get
∴ hCotΘ = (h + a) Cot(Θ + Φ)
⇒ CotΘ / Cot(Θ + Φ) = (h + a) / h
⇒ CotΘ / Cot(Θ + Φ) = h / h + a / h
⇒ CotΘ / Cot(Θ + Φ) = 1 + a / h
⇒ (CotΘ / Cot(Θ + Φ)) - 1 = a / h
⇒ (CotΘ - Cot(Θ + Φ) ) / Cot(Θ + Φ) = a / h
⇒ h = aCot (Θ + Φ) / CotΘ - Cot(Θ + Φ)
After converting Cot into Sin and Cos, We will get
⇒ h = aSinΘCos(Θ + Φ)/SinΦ
- A 6 ft-tall man finds that the angle of elevation of the top of a 24 ft-high pillar and the angle of depression its base are complementary angles. Then the distance between pillar and man is ?
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Let us draw a figure below from the given question.
Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively.
Here ∠ ACE = ∝ ∠ CBD = 90° - ∝
Correct Option: C
Let us draw a figure below from the given question.
Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively.
Here ∠ ACE = ∝ ∠ CBD = 90° - ∝
Now from right triangle CDB,
BD = 6 cot (90° - ∝) = 6 tan ∝
∴ tan ∝ = BD/6 ...............(i)
From right triangle ACE,
tan ∝ =18/EC = 18/BD ...............(ii) (∵ EC = BD)
Now from (i) and (ii) we get,
BD/6 = 18/BD
⇒ BD2 = 18 x 6
∴ BD = 6√3 ft.
Hence, distance between pillar and man = 6√3 ft
- Two ships leave a port at the same instant. One sails at 30km/hr in the direction N 32° E while the other sails at 20 km/hr in the direction S 58° E. After two hours the ships are distant from each other by.
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Let us draw a figure below as per given question.
Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.
Correct Option: C
Let us draw a figure below as per given question.
Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.
Now, ∠NOA = 32° and ∠SOB = 58° ,
Then, ∠AOB = 180° - (32° + 58°) = 90°
Since ΔAOB is a right triangle in which OA = 2 x 30 = 60 km and OB = 2 x 20 = 40 km
Since, AB = √OA2 + OB2
= √(60)2 + (40)2
= √5200
= 20√13
- ABC is a triangular park with AB = AC = 100 meters. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower of A and B are cot-1 (3.2) and cosec-1(2.6) respectively. The height of the tower in meters is ?
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Let us draw a figure as per given question.
Given ∝ = cot-1 (3.2) and β = cosec-1(2.6)
In triangle ΔPAD, AD = h cot ∝
In triangle ΔPBD, BD = h cot β
Correct Option: B
Let us draw a figure below as per given question.
Given ∝ = cot-1 (3.2) and β = cosec-1(2.6)
In triangle ΔPAD, AD = h cot ∝
In triangle ΔPBD, BD = h cot β
In triangle Δ ABD,
AB2 = AD2 + BD2
= h2(cot2∝ + cot2 β )
⇒ 1002 = h2{ cot2 ∝ + (cosec2 β - 1) }
⇒ 1002= h2 { (3.2)2 + (2.6)2 - 1} = 16h2
⇒ h = 25 m.