Two posts are x meter apart and the height of one is double that of the other. If from the midpoint of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in meter) of the shorter post is :

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 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 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 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 ?AOB 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

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.