Test Level 2: Trigonometry - 2 - CAT MCQ

Gievn: cosθ = 5/21 sinθ

Dividing both numerator and denominator by sinθ, 

Let AB be the vertical pole and AD be the shadow when sun's altitude is 60°.

∴ ∠BDA = 60°
Similarly, ∠BCA = 30°
CD = 25 m (Given)
In ΔADB,
tan 60° = AB

In ΔBDE,

In the figure, BE = CD = 10000 ft
∠BAE = 60° and ∠CAD = 30°
In triangle ABE,

In triangle ACD,

Distance travelled in one minute = ED = BC = AC – AB

4 cos² A + 2 (1 - cos² A) = 3
4 cos² A + 2 - 2cos²A = 3
2 + 2 cos²A = 3
2(1 + cos² A) = 3
1 + cos² A = 3/2

A = 45°

AC = AP tan α
⇒ (1/2) AB = n AB tan α

Length of the ladder will remain the same.
∴ AC = ED

Now,

[Using formula, cos C - cos D = 2 sin]

[since, sin(-θ) = - sinθ, cos (90° - 50°) = sin 50°]

[Using formula, sin C - sin D = 2 sin ]

= 2 cos 30° =√3

m² - n² = (m + n)(m - n) = 2 tan α × 2 sin α = 4 tan α sin α
Now, going by options:

Given height of the observer, DE = 1.6 m
Hence, BC = 1.6 m
AC is the height of the tower.
Let AB = h.
Given distance between the observer and the tower, CD = 45 m
⇒ BE = 45 m
In right ΔABE,

tan 30° = (h/45)

45 = h√3
h = 45 ÷ 1.732 = 25.98 m
Thus, height of tower = AC = AB + BC = 25.98 m + 1.6 m = 27.58 m
Hence, answer option 3 is correct.

AD is the tower and BC = 20 m
In triangle ACD,

AD = AE + ED
ED = BC = 20 m

AE + 20 = √3 CD . ....,. (1)
In triangle ABE,

Since, BE = CD
⇒ CD = √3 AE
On putting the value in equation 1, we get
AE + 20 = 3AE
⇒ 2AE = 20
⇒ AE = 10 m
So, AD = 10 + 20 = 30 m
∴ The height of the tower is 30 m.

n = sinθ - cosθ

Hence, 120 = h + 30 + h
⇒ h = 45 m
∴ Height of the vertical poles = 45 m