# Trigonometry & Height and Distances | Set-2

**Question 1:** Find the maximum value of (2sinθ + 3 cosθ)

**Solution :** Max value of (asinθ + bcosθ) = √(a^{2} + b^{2}))

So, max value = √(2^{2} + 3^{2})

= √(4 + 9)

= √13

**Question 2:** Find the minimum value of the 4tan^{2}θ + 9cot^{2}θ

**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 7sin^{2}θ + 3cos^{2}θ = 4, then the value of tanθ is

**Solution :** 7sin^{2}θ + 3cos^{2}θ = 4

7sin^{2}θ + 3(1-sin^{2}θ) = 4

7sin^{2}θ + 3 – 3sin^{2}θ = 4

4sin^{2}θ = 1

sin^{2}θ = 1/4

sinθ = 1/2

So, θ = 30^{o}

then tanθ = 1/√3

**Question 4:** is equal to

**Solution :** It can be solved by rationalization

Use the property:-

(1-sin^{2}θ = cos^{2}θ)

= 2/cosθ = 2 secθ

**Question 5:** If cosα = a cosβ and sinα = b sinβ, then the value of sin^{2}β in terms of a and b is:

**Solution :** On squaring both sides

cos^{2}α = a^{2} cos^{2}β

=> 1 – sin^{2}α = a^{2}(1 – sin^{2}β) ……..(1)

Again, sinα = b sinβ

On squaring both sides

sin^{2}α = b^{2} sin^{2}β

Put the value of sin^{2}α in (1)

1 – b^{2} sin^{2}β = a^{2} – a^{2}sin^{2}β)

a^{2} – 1 = a^{2}sin^{2}β – b^{2} sin^{2}β

a^{2} – 1 = sin^{2}β(a^{2} – b^{2})

sin^{2}β = a^{2} – 1 / (a^{2} – b^{2})

**Question 6:** The value of tan1^{o} tan2^{o} tan3^{o}…….tan89^{o} is

**Solution :** Arrange such that A + B = 90^{o}

(tan1^{o}tan89^{o})(tan2^{o}tan88^{o})…..(tan44^{o}tan46^{o})tan45^{o}

= 1x1x1……1×1

= 1

**Question 7:** If sin 2A = cos(A – 15^{o}), where 2A is an acute angle then the value of A is

**Solution :** sin 2A = cos(A – 15^{o})

2A + A – 15^{o} = 90^{o}

3A = 105^{o}

A = 35^{o}

**Question 8:**

then the value of sinθ is

**Solution :** By componendo and dividendo

tanθ/cotθ = 3/1

=> sin^{2}θ = 3cos^{2}θ

=> sin^{2}θ = 3(1 – sin^{2}θ)

=> 4sin^{2}θ = 3

=> sin^{2}θ = 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 30^{o}

=> θ/2 = 30^{o}

=> θ = 60^{o}

then cosθ = cos60^{o} = 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 a^{2} + b^{2} + c^{2} = ab + bc + ca, then the value of (sin^{2}A + sin^{2}B + sin^{2}C) is

**Solution :** Acc. to question

a^{2} + b^{2} + c^{2} – ab – bc – ca = 0

=>2a^{2} + 2b^{2} + 2c^{2} – 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 = 60^{2}

So, sin^{2}60 + sin^{2}60 + sin^{2}60

= 3(√3/2)^{2}

= 9/4

**Question 11:** The angle of elevation of an aeroplane from a point on the ground is 30^{o}. After flying for the 30 seconds, the elevation changes to 60^{o}. If the aeroplane is flying at height of 4500 metres, 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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