All questions of Trigonometry for Civil Engineering (CE) Exam

Writing the equation as :-

(log sin 1° - log cos 89°) + (log sin 2° - log cos 88°) + (log sin 3° - log cos 87°)………… + log tan 1°. log tan 89° + log tan 2°. log tan 88° + …….

=) (log sin 1° - log sin 1°) +(log sin 2° - log sin 2°)+……..+ log tan 1°cot 1° + log tan 2°cot 2°

=) log 1 = 0
 

 

Let BC be the height of the tower and DC be the height of the student.

3sinx + 4cosx ≥ -r

5(3/5sinx+4/5cosx) ≥ -r

35 = cosA => sinA =45

5(sinx cosA + sinA cosA) ≥ -r

5(sin(x + A)) ≥ -r

5sin (x + A) ≥ -r

-1 ≤ sin (angle) ≤ 1

5sin (x + A) ≥ -5

rmin = 5

We know that Sin²x + Cos²x = 1 for all values of x. 
If Sin x or Cos x is equal to –1 or 1, then Sin²⁰¹⁴x + Cos²⁰¹⁴x will be equal to 1.

Sin x is equal to –1 or 1 when x = –4.5π or –3.5π or –2.5π or –1.5π or –0.5π or 0.5π or 1.5π or 2.5π or 3.5π or 4.5π. 
Cosx is equal to –1 or 1 when x = –5π or –4π or –3π or –2π or –π or 0 or π or 2π or 3π or 4π or 5π.

For all other values of x, Sin²⁰¹⁴ x will be strictly lesser than Sin²x. 
For all other values of x, Cos²⁰¹⁴ x will be strictly lesser than Cos²x. 

We know that Sin²x + Cos²x is equal to 1. Hence, Sin²⁰¹⁴x + Cos²⁰¹⁴x will never be equal to 1 for all other values of x. Thus there are 21 values. 
Answer choice (C)

Let the hexagon ABCDEF be of side ‘a’. Line AD = 2a. Let towers at B and D be B’B and D’D respectively. 
From the given data we know that ∠B´AB = 30° and ∠D´AB = 45°. Keep in mind that the Towers B’B and D´D are not in the same plane as the hexagon.

Therefore, the answer is Option A.

Cos x – Sin x = √2 Sin x 

=> Cos x = Sin x + √2 Sin x 
=> Cos x = Sin x + √2 Sin x 
=> Sin x = Cosx/(√2+1) * Cos x 
=> Sin x = (√2−1)/(√2−1) * 1/(√2+1) * Cos x
=> Sin x = (√2−1)/((√2)²−(1)²)* Cos x
=> Sin x = (√2 - 1) Cos x
=> Sin x = √2 Cos x – Cos x
=> Sin x + Cos x = √2 Cos x

Hence, the correct answer is Option A.

Explanation : Cos A = 1 - Cos²A

=> Cos A = Sin²A

=> Cos²A = Sin⁴A

=> 1 – Sin²A = Sin⁴A

=> 1 = Sin4⁴A + Sin²A

=> 1³ = (Sin⁴A + Sin²A)³

=> 1 = Sin¹² A + Sin⁶A + 3Sin⁸A + 3Sin¹⁰A

=> Sin¹²A + Sin⁶A + 3Sin⁸A + 3Sin¹⁰A – 1 = 0

On comparing,

a = 1, b = 3 , c = 3 , d = 1

=  a+b/c+d

Hence, the answer is 1

The tops of two poles of height 30 m and 14 m are connected by a string. If the wire makes an angle of 30° with the horizontal.

Calculation: Let the length of the wire be h.

Height of pole 1 = 30 m AB = 30 - 14 = 16 m

In ΔABC, Sin30° = AB/AC ⇒ 1/2 = 16/h
⇒ h = 32 m

∴ The length of the wire is 32 m.

First find the distance of the diagonal d along the ground from corner to corner. Using Pythagorean theorem with sides 25 and 25, we get:

25² + 25² = d²

2 × 25² = d²

d = 25√ 2 .

Then to obtain the height h of the tree, use the tangent ratio with angle 60°.

tan 60° = x / (25√ 2 )

√ 3  = x / (25√ 2 )

x = 25√ 2  × √ 3  = 25√ 6 

We know that,
h² = p² + b² Given, p and b are positive integer, so h2 will be sum of two perfect squares.

We see 
a) 7² + 5² = 74
b) 6² + 4² = 52
c) 3² + 2² = 13
d) Can’t be expressed as a sum of two perfect squares

Therefore the answer is Option D.

Let’s say sin x – sin y = k. So we intend to find the range of k.
We have cos x + cos y = 1

So, k² + 1² = (sin x - sin y )² + (cos x + cosy )²
k² + 1² = (sin x)² + (sin y)² – 2(sin x)(sin y) + (cos x)² + (cos y)² + 2(cos x)(cos y)
k² + 1²= sin² x + cos² x + sin² y + cos² y + 2(cos x)(cos y) – 2(sin x)(sin y)
k² + 1² = 1 + 1 + 2(cos x)(cos y) – 2(sin x)(sin y) {Because, sin² x + cos² x = 1}
k² + 1²= 2 + 2[(cos x)(cos y) – (sin x)(sin y)] {Because, (cos x)(cos y) – (sin x)(sin y) = cos(x + y)}
k² = 1 + 2(cos(x + y))

The maximum value that a cos function can take is 1, and the minimum value is -1.

Therefore the maximum value that k² can take is 1+2(1) = 3,
and the minimum value that k² can take is 1+2(-1) = -1,
This is impossible, k² is always non-negative.
The minimum value that k² can take is 0.

Therefore (k²)_max = 3 and (k²)_min = 0.

To get the range of k, we concentrate on (k²)_max.

(k²)_max = 3, implies that the extreme values of k are:

k = +√3 or -√3
So, the values of k range between +√3 and -√3, both +√3 and -√3 inclusive.

k = sin x – sin y, Remember!!

Therefore, Range of sin x – sin y = [-√3, √3]