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Maths 4 Marks Questions

P-2 Trigonometry · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 28 questions.

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4 Marks Questions

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Question 14 Marks
If $\sec \theta=\frac{13}{12}$ find the values of other trigonometric ratios.
Answer
We know that,
$\sec ^2 \theta=1+\tan ^2 \theta$
$\Rightarrow \tan ^2 \theta=\sec ^2 \theta-1$
$\Rightarrow \tan ^2 \theta=\left(\frac{13}{12}\right)^2-1$
$\Rightarrow \tan ^2 \theta=\frac{169}{144}-1=\frac{25}{144}$
$\Rightarrow \tan \theta=\frac{5}{12} \ldots$
Also,
$\cos \theta=\frac{1}{\sec \theta}$
$\Rightarrow \cos \theta=\frac{12}{13} \ldots$
Now, using
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \frac{5}{12}=\frac{\sin \theta}{\frac{12}{13}}$
$\Rightarrow \sin \theta=\frac{5}{12} \times \frac{12}{13}$
$\Rightarrow \sin \theta=\frac{5}{13} \ldots[3]$
Also,
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{13}{5} \ldots$
$\cot \theta=\frac{1}{\tan \theta}=\frac{12}{5}$
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Question 24 Marks
If tan θ = 2, find the values of other trigonometric ratios.
Answer
We know that,
$\sec ^2 \theta=1+\tan ^2 \theta$
$\Rightarrow \sec ^2 \theta=1+(2)^2$
$\Rightarrow \sec ^2 \theta=5$
$\Rightarrow \sec \theta=\sqrt{ } 5 \ldots[1]$
Also,
$\cos \theta=\frac{1}{\sec \theta}$
$\Rightarrow \cos \theta=\frac{1}{\sqrt{5}}$
Now, using
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow 2=\frac{\sin \theta}{\frac{1}{\sqrt{5}}}$
$\Rightarrow \sin \theta=2 \times \frac{1}{\sqrt{5}}$
$\Rightarrow \sin \theta=\frac{2}{\sqrt{5}} \ldots[3]$
Also,
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{\sqrt{5}}{2}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{2}$
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Question 34 Marks
While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)
Answer

Let the initial position of plain be P and after landing it position be R.
Now,
Angle of depression while landing, ∠CPR = 20°
Also, ∠ CPR = ∠PRQ (say θ) = 20° [Alternate Angles]
Now,
Speed of plane = 200 km / hr
$=200 \times \frac{1000}{3600}=\frac{500}{9} ms^{-1}$
Time taken for landing $=54$ seconds Distance travelled in landing $=P R$
Also, distance $=$ speed $\times$ time
$\Rightarrow PR =\frac{500}{9} \times 54=3000 m$
Now, In $\triangle P Q R$
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{ PQ }{ PR }$
$\Rightarrow \sin 20^{\circ}=\frac{ PQ }{3000}$
⇒ PQ = 3000 × sin 20°
⇒ PQ = 3000(0.342)
⇒ PQ = 1026 m
So, Plane was at a height of 1026 m at the start of landing.
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Question 44 Marks
A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20m. If the platform is 2m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
Answer

Let AB be the ladder, i.e. AB = 20 m and PQRS be the platform.
In the above figure, clearly
PQ = RS = CD = height of platform = 2 m
Maximum height ladder can reach = BC + CD
Now,
Maximum value of ∠BAC = 70°
In right-angled triangle ABC,
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{ BC }{ AB }$
$\Rightarrow \sin 70^{\circ}=\frac{ BC }{20} $
$ \Rightarrow 0.94=\frac{ BC }{20}$
⇒ BC = 18.8 m
Maximum height ladder can reach = BC + CD
= 18.8 + 2 = 20.8 meters
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Question 54 Marks
Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?
Answer

Let AB and CD be two building, with
AB = 12 m
And angle of elevation from top of AB to top of CD = ∠CAP = 30°
Width of road = BD = 15 m
Clearly, ABDP is a rectangle
With
AB = PD = 12 m
BD = AP = 15 m
And APC is a right-angled triangle, In ∆APC
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{ CP }{ AP } $
$ \Rightarrow \tan 30^{\circ}=\frac{ CP }{15}$
$ \Rightarrow \frac{1}{\sqrt{3}}=\frac{ CP }{15} $
$ \Rightarrow CP =\frac{15}{\sqrt{3}}=\frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{15 \sqrt{3}}{3}=5 \sqrt{3}$
⇒ CP = 5√3 m
Also,
CD = CP + PD = (5√3 + 12) m
Hence, height of other building is (12 + 5√3 m).
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Question 64 Marks
From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.
Answer

Let PQ be a light house of height 80 cm such that PQ = 100 m
And R be a ship.
Angle of depression from P to ship R = ∠BPR = 30°
Also, ∠PRQ(say θ) = ∠BPR = 30° [Alternate Angles]
Clearly, PQR is a right-angled triangle.
Now, In ∆PQR
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{ PQ }{ QR } $
$ \Rightarrow \tan 30^{\circ}=\frac{100}{ QR } $
$ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{ QR }$
⇒ QR = 100√3 m
Hence, Ship is 100√3 m away from the light house.
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Question 74 Marks
A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Answer

Let 'R' be the person, standing 48 m away from a building PQ,
Angle of elevation, ∠PRQ = θ = 30°
Clearly, ∆ABC is a right-angled triangle, in which
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{P Q}{Q R} $
$ \Rightarrow \tan 30^{\circ}=\frac{B C}{48} $
$ \Rightarrow \frac{1}{\sqrt{3}}=\frac{B C}{48}$
$ \Rightarrow B C=\frac{48}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{48 \sqrt{3}}{3}=16 \sqrt{3} m$
Therefore, Height of church is 16√3 m.
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Question 84 Marks
If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
Answer
5secθ - 12cosecθ = 0
⇒ 5secθ = 12cosecθ
$\Rightarrow \frac{\sec \theta}{\operatorname{cosec} \theta}=\frac{12}{5}$
$\Rightarrow \frac{\frac{1}{\cos \theta}}{\frac{1}{\sin \theta}}=\frac{12}{5}$
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{12}{5}$
As $\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \tan \theta=\frac{12}{5}$
Also, We know that,
$\sec ^2 \theta=1+\tan ^2 \theta$
$\Rightarrow \sec ^2 \theta=1+\left(\frac{12}{5}\right)^2$
$\Rightarrow \sec ^2 \theta=1+\frac{144}{25}$
$\Rightarrow \sec ^2 \theta=\frac{169}{25}$
$\Rightarrow \sec \theta=\frac{13}{5}$
Also,
$\Rightarrow \cos \theta=\frac{1}{\sec \theta}$
$\Rightarrow \cos \theta=\frac{5}{13}$
Now, again using
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \frac{12}{5}=\frac{\sin \theta}{\frac{5}{13}}$
$\Rightarrow \sin \theta=\frac{12}{5} \times \frac{5}{13}$
$\Rightarrow \sin \theta=\frac{12}{13}$
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Question 94 Marks
Roshani saw an eagle on the top of a tree at an angle of elevation of 61°, while she was standing at the door of her
house. She went on the terrace of the house so that she could see it clearly. The terrace was at a height of 4m. While
observing the eagle from there the angle of elevation was 52°. At what height from the ground was the eagle ?
(Find the answer correct upto nearest integer)
(tan 61° = 1.80, tan 52° = 1.28, tan 29° = 0.55, tan 38° = 0.78)
Answer
In figure $6.12, \mathrm{PQ}$ is the house and $\mathrm{SR}$ is the tree. The eagle is at $\mathrm{R}$.
Draw seg QT $\perp$ seg RS.
$\therefore \square$ TSPQ is a rectangle.
Let $\mathrm{SP}=x$ and $\mathrm{TR}=y$
$ \text { Now in } \triangle \mathrm{RSP}, \angle \mathrm{PRS}=90^{\circ}-61^{\circ}=29^{\circ}$
$\text { and in } \triangle \mathrm{RTQ}, \angle \mathrm{QRT}=90^{\circ}-52^{\circ}=38^{\circ}$
$\therefore \tan \angle \mathrm{PRS}=\tan 29^{\circ}=\frac{\mathrm{SP}}{\mathrm{RS}}$
$\therefore 0.55=\frac{x}{y+4}$
$\therefore x=0.55(y+4) \ldots \ldots . . \text { (I) } $
Similarly, $\tan \angle \mathrm{QRT}=\frac{\mathrm{TQ}}{\mathrm{RT}}$
$ \left.\therefore \tan 38^{\circ}=\frac{x}{y} \ldots \ldots \ldots . . . \mathrm{SP}=\mathrm{TQ}=x\right]$
$\therefore 0.78=\frac{x}{y}$
$\therefore x=0.78 y \ldots \ldots \ldots \text { (II) }$
$\therefore 0.78 y=0.55(y+4) \ldots \ldots \ldots \text { from (I) and (II) }$
$\therefore 78 y=55(y+4)$
$\therefore 78 y=55 y+220$
$\therefore 23 y=220$
$\therefore y=9.565=10 \text { (upto nearest integer) }$
$\therefore \mathrm{RS}=y+4=10+4=14 $
from (I) and (II)
$\therefore$ the eagle was at a height of 14 metre from the ground.In figure $6.12, \mathrm{PQ}$ is the house and $\mathrm{SR}$ is the tree. The eagle is at $\mathrm{R}$.
Draw seg QT $\perp$ seg RS.
$\therefore \square$ TSPQ is a rectangle.
Let $\mathrm{SP}=x$ and $\mathrm{TR}=y$
$ \text { Now in } \triangle \mathrm{RSP}, \angle \mathrm{PRS}=90^{\circ}-61^{\circ}=29^{\circ}$
$\text { and in } \triangle \mathrm{RTQ}, \angle \mathrm{QRT}=90^{\circ}-52^{\circ}=38^{\circ}$
$\therefore \tan \angle \mathrm{PRS}=\tan 29^{\circ}=\frac{\mathrm{SP}}{\mathrm{RS}}$
$\therefore 0.55=\frac{x}{y+4}$
$\therefore x=0.55(y+4) \ldots \ldots . . \text { (I) } $
Similarly, $\tan \angle \mathrm{QRT}=\frac{\mathrm{TQ}}{\mathrm{RT}}$
$ \left.\therefore \tan 38^{\circ}=\frac{x}{y} \ldots \ldots \ldots . . . \mathrm{SP}=\mathrm{TQ}=x\right]$
$\therefore 0.78=\frac{x}{y}$
$\therefore x=0.78 y \ldots \ldots \ldots \text { (II) }$
$\therefore 0.78 y=0.55(y+4) \ldots \ldots \ldots \text { from (I) and (II) }$
$\therefore 78 y=55(y+4)$
$\therefore 78 y=55 y+220$
$\therefore 23 y=220$
$\therefore y=9.565=10 \text { (upto nearest integer) }$
$\therefore \mathrm{RS}=y+4=10+4=14 $ from (I) and (II)
$\therefore$ the eagle was at a height of 14 metre from the ground.
Image
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Question 104 Marks
To find the width of the river, a man observes the top of a tower on the opposite bank making an angle of elevation of $61^{\circ}$. When he moves $50 \mathrm{~m}$ backword from bank and observes the same top of the tower, his line of vision makes an angle of elevation of $35^{\circ}$. Find the height of the tower and width of the river. $\left(\tan 61^{\circ}=1.8, \tan 35^{\circ}=0.7\right)$
Answer
seg $A B$ shows the tower on the opposite bank. ' $A$ ' is the top of the tower and seg BC shows the width of the river. Let ' $h$ ' be the height of the tower and ' $x$ ' be the width of the river.
from figure, $\tan 61^{\circ}=\frac{h}{x}$
$\therefore 1.8  =\frac{h}{x}$
$h  =1.8 \times x$
$10 h  =18 x \ldots \ldots $
(I)..... multipling by $10$
In right angled $\triangle \mathrm{ABD}$,
$ \tan 35=\frac{h}{x+50}$
$\qquad 0.7=\frac{h}{x+50}$
$\therefore \quad h=0.7(x+50)$
$\therefore \quad 10 h=7(x+50) \ldots . . . . \text { (II) }$
$\therefore \text { from equations (I) and (II) },$
$\quad 18 x=7(x+50)$
$\therefore \quad 18 x=7 x+350$
$\therefore \quad 11 x=350$
$\therefore \quad x=\frac{350}{11}=31.82$
$\text { Now, } h=1.8 x=1.8 \times 31.82$
$\quad \quad=57.28 \mathrm{~m} . $
$\therefore$ width of the river $=31.82 \mathrm{~m}$ and height of tower $=57.28 \mathrm{~m}$Image
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Question 114 Marks
From a point on the roof of a house, 11 m high, it is observed that the angles of depression of the top and foot of a lamp post are 30 and 60 respectively. What is the height of the lamp post?
Answer
7.33 m
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Question 124 Marks
A ship of height 24 m is sighted from a lighthouse. From the top of the lighthouse, the angle of depression to the top of the mast and base of the ship is 30 and 45 respectively. How far is the ship from the lighthouse? $(\sqrt{3}=1.73)$
Answer
56.76 m
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Question 134 Marks
A straight road leads to the foot of a tower of height 50 m. From the top of the tower, the angle of depression of two cars standing on the road are 30 and 60. What is the distance between the two cars?
Answer
$\frac{100}{\sqrt{3}}$
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Question 144 Marks
From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be 60. If the lighthouse is 90 m, then find how far is that ship from the lighthouse? $(\sqrt{3}=1.73)$
Answer
51.9 m
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Question 174 Marks
While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)
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Question 184 Marks
A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20m. If the platform is 2m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
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Question 194 Marks
Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?
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Question 204 Marks
From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.
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Question 214 Marks
A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
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Question 234 Marks
Roshani saw an eagle on the top of a tree at an angle of elevation of 61°, while she was standing at the door of her
house. She went on the terrace of the house so that she could see it clearly. The terrace was at a height of 4m. While
observing the eagle from there the angle of elevation was 52°. At what height from the ground was the eagle ?
(Find the answer correct upto nearest integer)
(tan 61° = 1.80, tan 52° = 1.28, tan 29° = 0.55, tan 38° = 0.78)
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Question 244 Marks
To find the width of the river, a man observes the top of a tower on the opposite bank making an angle of elevation of $61^{\circ}$. When he moves $50 \mathrm{~m}$ backword from bank and observes the same top of the tower, his line of vision makes an angle of elevation of $35^{\circ}$. Find the height of the tower and width of the river. $\left(\tan 61^{\circ}=1.8, \tan 35^{\circ}=0.7\right)$
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Question 254 Marks
From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be 60. If the lighthouse is 90 m, then find how far is that ship from the lighthouse? $(\sqrt{3}=1.73)$
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Question 264 Marks
From a point on the roof of a house, 11 m high, it is observed that the angles of depression of the top and foot of a lamp post are 30 and 60 respectively. What is the height of the lamp post?
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Question 274 Marks
A straight road leads to the foot of a tower of height 50 m. From the top of the tower, the angle of depression of two cars standing on the road are 30 and 60. What is the distance between the two cars?
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Question 284 Marks
A ship of height 24 m is sighted from a lighthouse. From the top of the lighthouse, the angle of depression to the top of the mast and base of the ship is 30 and 45 respectively. How far is the ship from the lighthouse? $(\sqrt{3}=1.73)$
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