Test Level 2: Trigonometry - 1 - CAT MCQ

(cosθ cos(90 - θ) secθ) ÷ (tanθ sin (90 - θ))
= cosθ sinθ secθ ÷ tanθ cosθ
= cosθ sinθ secθ ÷ cosθ sinθ secθ
= 1 

a = x sinθ+ y cosθ
b = y sinθ- x cosθ
y² - a² = y² - (x sinθ+ y cosθ)²
= y² - x² sin²θ - y² cos2θ - 2xy sinθ cosθ
= y² sin²θ - x² sin²θ - 2xy sinθ cosθ
= y² sin²θ + x² cos²θ - x² - 2xy sinθ cosθ
= (y sinθ - x cosθ)² - x²
= b² - x²

Given: sin (π cos x) = cos (π sin x)

In ΔABC
AB/BC = tan 60°
AB = √3b units
In Δ ABD
AB/BD = tan 30°
AB = a/√3 units
Now AB × AB = √3b × a/√3 units² 
AB² = ab units²
AB = √ab units

Distance travelled in 10 minutes = CD = x metres

Solving (i) and (ii), we get

CB + x = 5CB
x = 4CB
or, CB = x/4 metres
Total time taken to cover a distance of x/4 metres 

Multiply and divide by 2,

The componendo - dividendo method state that
Using above theorem in equation (1) we get

The tower makes equal angles at the vertices of the triangle.
Therefore, foot of the tower is at the circumcentre.
From ΔOAP, we have tanα = OP/OA
⇒ OP = OA tan α
⇒ OP = R tan α

In the diagram, BC = 30 m = DE, ∠ABD = 45° and ∠EBD = 60°.

Let AD = x m and AE = (x + 30) m.
In right-angled ΔABD,
tan 45° = AD/BD
⇒ 1 = x/BD ⇒ BD = x m … (1)
In right-angled ΔEBD,
tan 60° = DE/BD

From equations (1) and (2), we get

∴ Height of the tower = AE = (x + 30) m
= (10√ + 30) m
= (10 × 1.732 + 30) m = 47.32 m