Two ships leave a port at the same instant. One sails at 30km/hr in the direction N 32° E while the other sails at 20 km/hr in the direction S 58° E. After two hours the ships are distant from each other by.

Let us draw a figure below from the given question.
Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively.
Here ? ACE = ? ? CBD = 90° - ?
Now from right triangle CDB,
BD = 6 cot (90° - ?) = 6 tan ?
? tan ? = BD/6 ...............(i)
From right triangle ACE,
tan ? =18/EC = 18/BD ...............(ii) (? EC = BD)
Now from (i) and (ii) we get,
BD/6 = 18/BD
? BD² = 18 x 6
? BD = 6?3 ft.
Hence, distance between pillar and man = 6?3 ft

Let us draw a figure as per given question.
Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that ?OAP = ? and ?PAQ = ?
In ?OAP use the formula
Tan? = P/B = Perpendicular distance / Base distance
? Tan? = OP/OA
? OA = OP Cot?
Put the value of OP in above equation, we will get
? OA = hCot? ................ (1)
Now from triangle ?OAQ,
? Tan(? + ?) = OQ/OA
? OA = OQ Cot(? + ?)
? OA = (h + a) Cot(? + ?) .............(2)
from equation (1) and (2), We will get
? hCot? = (h + a) Cot(? + ?)
? Cot? / Cot(? + ?) = (h + a) / h
? Cot? / Cot(? + ?) = h / h + a / h
? Cot? / Cot(? + ?) = 1 + a / h
? (Cot? / Cot(? + ?)) - 1 = a / h
? (Cot? - Cot(? + ?) ) / Cot(? + ?) = a / h
? h = aCot (? + ?) / Cot? - Cot(? + ?)
After converting Cot into Sin and Cos, We will get
? h = aSin?Cos(? + ?)/Sin?

Let us draw a figure below as per given question.
Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ?ADC = 30° ?ACB = 45° and BC = x meter (say)
Now from right triangle ABC,
tan 45° = h/x ? 1 = h/x
? x = h
Again from right triangle ABD,
tan 30° = h/( 200 + x )
? 1/?3 = h/( 200 + x )
Since x = h , we will get.
? 1/?3 = h/(200 + h )
? (200 + h ) = h X ?3
? h X ?3 - h = 200
? h x 1.732 - h = 200
? h(1.732 - 1) = 200
? h = 200/0.732 = 273.2 m = 273 m

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

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 ?,
? In triangle ? PCD, tan 30° = h/a
i.e h/a = 1/?3 .............. (1)
And in triangle ? PBD,
Since PD = h and BD = ?(AB² + AD)² = a?2.
Put the value of PD and BD , we will get
tan ? = PD/BD = h/(a?2)
put the value of h/a from equation (1)
tan ? = 1/?2 x 1/?3
tan ? = 1/?6

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 ?OAE, Sin ?/2 = OA/OE
? OE = r cosec 1/2 ?
In ?OEL, height of the center of the balloon = h = OE sin ? = r Cosec ?/2 Sin ?.

Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that
?OAP = ? and ?PAQ = ?
In ?OAP and ?OAQ
OA = OP cot ? = h cot?
and OA = OQ cot ( ? + ? ) = (h + a) cot ( ? + ? )
? h cot? = (h + a) cot ( ? + ? )
? h = a cot ( ? + ? ) / cot? - cot ( ? + ? )