Test Level 3: Trigonometry Solved MCQs CAT

Let BC = BD = x
∠CPM = α and ∠MPD = β

 

Also, BM = AP = h
∴ MC = x - h
and DM = x + h
Now, in CPM;
PM = MC cot α = (x - h)cot α …(i)
In ΔDPM;
PM = DM cot β = (x + h) cot β …(ii)
From eq. (i) and (ii), we get:
(x - h) cot α = (x + h) cot β

Distance travelled in 10 sec = AB = PQ = OQ - OP

If h₁, h₂, h₃, are altitudes, then h₁ = c sin B; h₂ = a sin C; h₃ = b sin A


Given sin A, sin B sin C..... are in A.P

7 cos²x + sin x cos x - 3 = 0
⇒ 7 + tan x - 3 sec²x = 0 (dividing both sides by cos2x)
⇒ 7 + tan x - 3(1 + tan²x) = 0
⇒ 3 tan²x - tan x - 4 = 0

n, K ∈ I

xcotθ – ycosecθ = a
Squaring both sides, we get
x²cot²θ + y²cosec²θ – 2.xy.cotθ.cosecθ = a² --- (1)
xcosecθ – ycotθ = b
Squaring both sides, we get
x²cosec²θ + y²cot²θ – 2.xy.cosecθ.cotθ = b² --- (2)
Subtracting (2) from (1), we get
x²cot2θ + y²cosec²θ – 2.xy.cotθ.cosecθ - (x²cosec²θ + y²cot2θ – 2.xy.cosecθ.cotθ) = a² - b²
x²cot²θ + y²cosec²θ - x²cosec²θ - y²cot²θ = a² - b²
cosec²θ(y² – x²) – cot²θ(y² – x²) = a² – b²
(y² – x²)(cosec²θ – cot²θ) = a² – b²
(y² – x²) = a² – b²
Squaring both sides,
(y² – x²)² = (a² - b²)²

cot(p + q – r + s) = 0
cot(p + q – r + s) = cot 90°
Therefore, (p + q – r + s) = 90° --- (1)
Now, sin(p – q) = 1/2
sin(p – q) = sin 30°
So,(p – q) = 30° --- (2)
Also, cosec(q + s) = √2
cosec(q + s) = cosec 45°
(q + s) = 45° --- (3)
And sec(q + r) = 1
sec(q + r) = sec 0°
(q + r) = 0° --- (4)
Now, adding equations (2) and (4),
p + r = 30° --- (5)
Now, putting (3) in (1),
So, p – r = 90° - 45° = 45° --- (6)
Now, subtracting equation (6) from equation (5),
2r = -15°
r = -7.5°
Now, putting r = -7.5° in equation (4), we get
q = 7.5°

After putting the values of x and y, the expression becomes:
(secθ – cosecθ)² – 4 – (tanθ – cotθ)² + 4/(cosθ sinθ)
= (sec²θ + cosec²θ - 2secθ cosecθ) - 4 - (tan²θ + cot²θ - 2tanθ cot θ) + 4/(cosθ sinθ)
= sec²θ + cosec²θ - 2secθ cosecθ – 4 - tan²θ - cot²θ + 2tanθ cot θ + 4/(cosθ sinθ)

[As sec²θ - tan²θ = 1 and cosec²θ - cot²θ = 1]

(i) sec⁴θ + sec²θ = 1 [Given]
sec²θ = 1 - sec⁴θ
1 + tan²θ = 1 – (1 + tan²θ)² [As sec2θ = 1 + tan²θ]
1 + tan²θ = 1 – [1 + tan⁴θ + 2tan²θ]
1 + tan²θ = 1 – 1 - tan⁴θ – 2tan²θ
1 + 3tan²θ + tan⁴θ = 0
3tan²θ + tan⁴θ = -1

Putting the values of p and q in this expression,

= tanθ + cotθ

(iii) o² – m² – n² + I²tan²θ
Putting the values of o, m and n in this expression,
= g²cosec²θ - g²cot²θ - I2sec²θ + I²tan²θ
= g²(cosec²θ - cot²θ) - I²(sec²θ - tan²θ)
= g² - I² [As cosec²θ - cot²θ = 1 and sec²θ - tan²θ = 1]

In the figure, let OP = n and OA = m
Then, length of the ladder 

In triangle OPB, 

n = (m + b) tan x
In triangle OAQ,

 n = m tan y - a
Equating the values of n, we get

In Δ ABE,


and in Δ BCD, 

From Eqs. (i) and (ii),

Put the value of tanθ, we get

In figure, the vertical line gives the position of rocket at different times. At B, the time is 1 second.
In triangle OBA,
OB = a + b and OA = 3

At t = 2 seconds, the position of the rocket is at C.
OC = 2a + b
In triangle OCA,

Solving for b and a,
a = 3 - √3 and b = 2√3 - 3
Hence, the expression for the distance becomes
s = (3 - √3) t + 2√3 - 3
In triangle ODA,

Total length = d sin30° + d sin 30° sin 60° + d sin 30° sin 60° sin 30° + ...

In figure, BC = 1 km 
In triangle APB,

In triangle PAC,

Equating the expressions, we get

From the figure, we find that
tan θ = 50/x (From triangle OMP) ... (1)
Also, tan 2θ = 300/x (From triangle OBM) ... (2)

⇒ 3(x² - 2500) = x²
⇒ x² = 3x² - 7500
⇒ 2x² = 7500
⇒ x² = 3750 = 625 × 6
⇒ x = 25√6
Hence, distance of the observer from the top of the pole is 25√6m.