Daily Practice Questions Questions for CAT with Answers PDF

cot (90 - θ) = tan θ
So, tan y = cot (90° - y)
Hence, cot (x + 27°) = cot (90° - y)
x + 27° = 90° - y
x + y = 90° - 27° = 63°

2 cos 3x - 1 = 0
⇒ cos 3x = 1/2 = cos 60°
The general solution of above equation is 

The solution in the interval (0, 2π) are

Hence, correct answer that match the given options are 20°, 100°

The value of cos 24° + cos 55° + cos 125° + cos 204° + cos 300° ?
We know that, cos(180° ± θ) = - cosθ
⇒ cos 24° + cos 55° + cos(180° - 55°) + cos(180° + 24°) + cos(360° - 60°)
= cos 24° + cos 55° - cos 55° - cos 24° + cos 60°
= cos 60° = 1/2

2 sin² x + 3 cos x = 0
⇒ 2(1 - cos₂ x) + 3 cos x = 0
⇒ 2 - 2 cos² x + 3 cos x = 0
⇒ 2 cos² x - 3 cos x - 2 = 0
⇒ 2 cos² x - 4 cos x + cos x - 2 = 0
⇒ 2 cos x(cos x - 2) + 1(cos x - 2) = 0
⇒ (cos x - 2)(2 cos x + 1) = 0
Either cos x - 2 = 0 or 2 cos x + 1 = 0
But cos x - 2 = 0
i.e. cos x = 2, which is not possible.
Now, from 2 cos x + 1 = 0, we get:
cos x = -1/2

Therefore, general solution of the equation is:

cos² 15° - cos² 30° + cos² 45° - cos² 60° + cos² 75° = (cos² 15° + cos² 75°) - (cos² 30° + cos² 60°) + cos² 45°
= [cos² 15° + cos² (90° - 15°)] - [cos² 30° + cos² (90° - 30°)] + cos² 45°
= [cos² 15° + sin² 15°] - [cos² 30° + sin² 30°] + cos² 45° [∵ cos (90° - θ) = sin θ]

Let AB and CD be the two pillars of equal height 'h' m.
If P is the point between them, then
∠APB = 60° and ∠CPD = 30°

Let BP = x m
⇒ PD = (150 - x) m
In ΔABP, tan 60° = 
⇒ h = √3x … (1)

⇒ 3x = 150 - x or 4x = 150
∴ x = 37.5 and h = √3x = (1.732)37.5 = 64.95 m

cosθ = 2 cos²θ

secθ = 2
secθ = sec 60°
θ = 60°
2 sec²θ – cos²θ + tan²θ = 2 sec² 60° - cos² 60° + tan² 60° = 

(∵ sin (90° - θ) = cos θ and cos (90° - θ) = sin θ)

Let A be the point of observation and C be the position of the cloud. Let DE be the surface of the lake and C' be the reflection of C in the lake.
∠CAB = 30°
And ∠C'AB = 60°
Let BC = h m

In right-angled ΔABC, BC/AB = tan 30°

In right-angled ΔABC',

h = 60 m
Height of cloud = 120 m

sin(x) + cos(x) = 1
Squaring both side,
(sin(x) + cos(x))² = 12
[sin²(x) + cos²(x)] + 2sin(x)cos(x) = 1 (Using identity: (a + b)² = a² + b² + 2ab)
1 + 2sin(x)cos(x) = 1
2sin(x)cos(x) = 0
sin(x)cos(x) = 0
From the last line above, either sine is zero or cosine is zero, so solutions are:
x = 0°, 90°, 180°, 270°
Checking the value of 'x' in the above equation,
x = 0°, 90° satisfies.
x = 180° and 270° does not satisfies.
Hence, total number of solution = 2