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Trigonometry & Height and Distances | Set-2

  • Last Updated : 16 Sep, 2021

Question 1: Find the maximum value of (2sinθ + 3 cosθ) 
Solution : Max value of (asinθ + bcosθ) = √(a2 + b2)) 
So, max value = √(22 + 32
= √(4 + 9) 
= √13 

Question 2: Find the minimum value of the 4tan2θ + 9cot2θ 
Solution : Minimum value = √ab 
Here a = 4 and b = 9 
Minimum value = √(4×9) 
= √36 
= 6 

Question 3: If θ be an acute angle and 7sin2θ + 3cos2θ = 4, then the value of tanθ is 
Solution : 7sin2θ + 3cos2θ = 4 
7sin2θ + 3(1-sin2θ) = 4 
7sin2θ + 3 – 3sin2θ = 4 
4sin2θ = 1 
sin2θ = 1/4 
sinθ = 1/2 
So, θ = 30o 
then tanθ = 1/√3 

Question 4: \sqrt{\frac{1+sin\theta }{1-sin\theta }}\:+\:\sqrt{\frac{1-sin\theta \:}{1+sin\theta \:}}  is equal to 
Solution : It can be solved by rationalization 
\frac{\left(\sqrt{1+sin\theta \:}\right)^2\:+\:\left(\sqrt{1-sin\theta \:\:}\right)^2}{\sqrt{1-sin^2\theta \:\:}}
Use the property:- 
(1-sin2θ = cos2θ) 
=\frac{1\:+sin\theta \:+\:1\:-sin\theta }{cos\theta }
= 2/cosθ = 2 secθ 

Question 5: If cosα = a cosβ and sinα = b sinβ, then the value of sin2β in terms of a and b is: 
Solution : On squaring both sides 
cos2α = a2 cos2β 
=> 1 – sin2α = a2(1 – sin2β) ……..(1) 
Again, sinα = b sinβ 
On squaring both sides 
sin2α = b2 sin2β 
Put the value of sin2α in (1) 
1 – b2 sin2β = a2 – a2sin2β) 
a2 – 1 = a2sin2β – b2 sin2β 
a2 – 1 = sin2β(a2 – b2
sin2β = a2 – 1 / (a2 – b2

Question 6: The value of tan1o tan2o tan3o…….tan89o is 
Solution : Arrange such that A + B = 90o 
(tan1otan89o)(tan2otan88o)…..(tan44otan46o)tan45o 
= 1x1x1……1×1 
= 1 

Question 7: If sin 2A = cos(A – 15o), where 2A is an acute angle then the value of A is 
Solution : sin 2A = cos(A – 15o
2A + A – 15o = 90o 
3A = 105o 
A = 35o 

Question 8: \frac{tan\theta \:+\:cot\theta }{tan\theta -cot\theta }=2\left(0\le \theta \le 90^o\right)
then the value of sinθ is 
Solution : By componendo and dividendo 
tanθ/cotθ = 3/1 
=> sin2θ = 3cos2θ 
=> sin2θ = 3(1 – sin2θ) 
=> 4sin2θ = 3 
=> sin2θ = 3/4 
sinθ = √/2 

Question 9: If tan (π/2 – θ/2) = √3 then the value of cosθ is 
Solution : tan (π/2 – θ/2) = √3 
=> tan (90 – θ/2) = √3 
=> cotθ/2 = √3 = cot 30o 
=> θ/2 = 30o 
=> θ = 60o 
then cosθ = cos60o = 1/2 

Question 10: a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a2 + b2 + c2 = ab + bc + ca, then the value of (sin2A + sin2B + sin2C) is 
Solution : Acc. to question 
a2 + b2 + c2 – ab – bc – ca = 0 
=>2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0 
=>(a-b)2 + (b-c)2 + (c-a)2 = 0 
=> a=b=c 
All three sides are equal then it is equilateral triangle. 
then ∠A = ∠B = ∠C = 602 
So, sin260 + sin260 + sin260 
= 3(√3/2)2 
= 9/4 

Question 11: The angle of elevation of an aeroplane from a point on the ground is 30o. After flying for the 30 seconds, the elevation changes to 60o. If the aeroplane is flying at height of 4500 meters, then the speed of the aeroplane in km/hr is 
Solution : 
 

In ∆ABE 
tan 60 = P/B = 4500/AB 
√3 = 4500/AB 
AB = 4500/√3 

In ∆ACD 
tan 30 = 4500/AC 
1/√3= 4500/AC 
AC = 4500√3 

Distance covered by plane in 30 seconds = AC – AB 
=4500√3 – 4500/√3 
= (13500 – 4500)/√3 
= 9000/√3 
= 3000√3 

Speed= Distance/time = 3000√3 / 30 
= 100√3 x 18/5 (to get km/hr) 
= 360√3 
Hence, speed of the plane is 360√3 km/hr

Question 12: There are two towers, one on each side of the road just opposite to each other. One tower is 54m high. From the top of this tower, the angles of the depression of the top and foot of the other tower are 30 and 60 respectively. The height of the other tower is: 
Solution : 
 

AB and CD are Towers. 
BD is the width of the road. 
AB = 54 m 
In ∆ AEC 
tan 30 = AE/EC = 1/√3 
=> AE : EC = 1 : √3 
In ∆ABD 
tan 60 = AB/BD 
√3 = AB/BD 
=> AB : BD = √3 : 1 
From diagram we know EB = CD and EC = BD 
Now, 

AB     :     BD     :     AE
             √3           1
√3     :      1
3      :     √3     :     1

CD = AB – AE = 3 – 1 = 2 units 

3 units of AB -> 54 m 
1 unit -> 18 
Then 2 units -> 36 m 
Hence, the height of the other tower is 36m.
 


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