The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 10 m. What was the height of the tree?

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 ( ? + ? )

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 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 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.