Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that
∠OAP = θ and ∠PAQ = ɸ
In ΔOAP and ∠OAQ

Correct Option: D

Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that
∠OAP = θ and ∠PAQ = ɸ
In ΔOAP and ∠OAQ
OA = OP cot θ = h cotθ
and OA = OQ cot ( θ + ɸ ) = (h + a) cot ( θ + ɸ )
∴ h cotθ = (h + a) cot ( θ + ɸ )
⇒ h = a cot ( θ + ɸ ) / cotθ - cot ( θ + ɸ )

Let O be the center of the balloon of radius r which subtend an angle ∝ at the eye of an observer at E.
If EA and EB are the tangents to the balloon,
then ∠ OEA = ∠ OEB = ∝/2

Correct Option: B

Let O be the centre of the balloon of radius r which subtend an angle ∝ at the eye of an observer at E .
If EA and EB are the tangents to the ballon,
then ∠ OEA = ∠ OEB = ∝/2
In triangle ΔOAE, Sin ∝/2 = OA/OE
∴ OE = r cosec 1/2 ∝
In ∠OEL, height of the center of the balloon = h = OE sin β = r Cosec ∝/2 Sin β.

Let us draw a figure below from given question.
Let a be the length of a side of square plot ABCD and h, the height of the pole standing at D. Since elevations of P from A or C is 30° and that from B is Θ,

Correct Option: B

Let a be the length of a side of square plot ABCD and h, the height of the pole standing at D. Since elevations of P from A or C is 30° and that from B is Θ,
∴ In triangle Δ PCD, tan 30° = h/a
i.e h/a = 1/√3 .............. (1)
And in triangle Δ PBD,
Since PD = h and BD = √(AB² + AD)² = a√2.
Put the value of PD and BD , we will get
tan Θ = PD/BD = h/(a√2)
put the value of h/a from equation (1)
tan Θ = 1/√2 x 1/√3
tan Θ = 1/√6

Let us draw a figure as per given question.
Given ∝ = cot^-1 (3.2) and β = cosec^-1(2.6)
In triangle ΔPAD, AD = h cot ∝
In triangle ΔPBD, BD = h cot β

Correct Option: B

Let us draw a figure below as per given question.
Given ∝ = cot^-1 (3.2) and β = cosec^-1(2.6)
In triangle ΔPAD, AD = h cot ∝
In triangle ΔPBD, BD = h cot β
In triangle Δ ABD,
AB² = AD² + BD²
= h²(cot²∝ + cot² β )
⇒ 100² = h²{ cot² ∝ + (cosec² β - 1) }
⇒ 100²= h² { (3.2)² + (2.6)² - 1} = 16h²
⇒ h = 25 m.

Let us draw a figure below as per given question.
Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.

Correct Option: C

Let us draw a figure below as per given question.
Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B.
Now, ∠NOA = 32° and ∠SOB = 58° ,
Then, ∠AOB = 180° - (32° + 58°) = 90°
Since ΔAOB is a right triangle in which OA = 2 x 30 = 60 km and OB = 2 x 20 = 40 km
Since, AB = √OA² + OB²
= √(60)² + (40)²
= √5200
= 20√13