Daily Practice Questions Questions for CAT with Answers PDF

Suppose OP is the building, and A and B are the two cars. Then, OP = 60, PAO = 60° and PBO = 30°
From the right triangle OBP, tan 30° =  60/OB
OB = 60 √m ...(i)
Also, from the right triangle AOP, tan 60° = 60/OA

OA = 60/√3 ....(ii)
Thus, from (i) and (ii),

If 10 sinθ = 6,
sinθ = 6/10 = 3/5

In a right-angled triangle PQR with ∠Q = 90°:

sin R = p/h , cos R = b/h, tan R = p/b, cosec R = h/p , sec R = h/b and cot R = b/p.
Here, p stands for perpendicular, b stands for base and h stands for hypotenuse.
Now, in ΔPQR, tan R = 1/√3 ……… (1)
Also, tan R = p/b……… (2)
On comparing (1) and (2), we get
p = 1 and b = √3
Using Pythagoras theorem:
h² = p² + b²
h = 
h = 2
Thus, p = 1 and h = 2
So, sin R = p/h  = 1/2.

Given: sin 30° = 1/2

In right ΔABC,
∠B = 90°, cot² A = 9/16
Taking square root on both sides, we get
cot A = 3/4… (1)

From (1) and (2), we get
b/p = 3/4
⇒ b = 3k and p = 4k
Now, using Pythagoras theorem,
AC² = AB² + BC²
h² = p² + b²
= (4k)² + (3k)²
= 16k² + 9k²
= 25k²
h = 5k
cos A = b/h = 3/5

Therefore, sec 30° = 2/√3 =  cosec 30° = 2, tan 30° = 1/√3 and cot 30° = √3
Hence, sec 30° tan 30° cot 30° cosec 30° 

1 + tan²θ = sec²θ,
1 + cot²θ = cosec²θ
and cotθ = 1/tanθ
Given:
m = 1 + tan²θ

When sinθ = 3/5, we can draw a triangle as shown in the figure.
sinθ = CB/AC = 3/5 (given)
cosθ = AB/AC = 4/5
tanθ = BC/AB = 3/4
cotθ = AB/BC = 4/3
cosθ + tanθ + cotθ = (4/5) + (3/4) + (4/3)
= (48 + 45 + 80)/60 = 173/60

or sec A = 2
or sec A = sec 60°
or A = 60°
Therefore, option (1) is correct.

The observer is at point 'D'.

We know that,
cos² x + sin² x = 1
Taking the third power,
(cos² x + sin² x)³ = cos⁶ x + sin⁶ x + 3 cos² x sin⁴x + 3 sin² x cos⁴ x = 1
i.e. cos⁶ x + sin⁶ x + 3 cos² x sin² x (sin² x + cos²x) = 1
cos⁶ x + sin⁶ x + 3 cos² x sin² x = 1

tan²θ (1 + cot²θ) (cosec²θ + sec²θ)

tan A = 3/4
⇒ sin A = 3/5 and cos A = 4/5

Let the length of the tower be CD.

CD = 120 m
∠DAC = 45° and DBC = 30°
In ΔADC,
DC/AC = tan 45°
120/AC = 1
AC = 120 m

CB = 120 × 1.732
CB = 207.84 m
Thus, distance between points A and B
= AC + CB = 120 + 207.84 = 327.84 m