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44. To find the height of an accessible object.

Let AC be the object, and let A any distance BC from its foot be measured.

At B let the angle of elevation
ABC be observed.
Suppose BC = 4, LABC = €

Then, we have

AC BL

+ BC = tan ABC or 4C – tan e. :: AC = a tan e, the height required. Ex. Let a = 200, and a = 30°.

1 200 13. Then AC = 200 tan 30o = 200. Tg = 3 45. To find the height of an inaccessible object.

YA At any point B in the

horizontal plane of the base let the angle of elevation ABC be observed.

Measure a convenient distance BD in the straight line

CB produced, and observe B C the angle of elevation ADC. Let BD = a, ZABC = 0, 2 ADB = $.

Then, Euc. I. 32, Z BAD = 0 – 0.
N AB sin ADB AB sin •
NOW BDsin BAD“ a sin(0 - 0)'

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sino

:: AB = a.

a

.

AC

....... (1). sin (0-°) Again, A = sin ABC = sin 0, :: AC = AB sin 0, or, from (1) AC = a sin o sin o.

sin (0 - 0) Ex. Let BD = 120, 0 = 60°, 6 = 45°.

Iued, AU 5.120 • sin (60° - 459)*

Then, AC = 120. sin 60° sin 450

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[ocr errors]

sin y

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= 120.13
0.73 - 1 =

= 60 (3 + 13). 46. To find the height of an inaccessible object when it is not convenient to measure any distance in a line with the base of the object.

Let a distance BD be measured in any direction in the same horizontal plane as BC, and let the angles ABC, ABD, ADB be observed. Let BD = a, LABC = a

ABD = B, L ADB = y. Then, Euc. I., 32, 2 BAD = 108° - (B + y).

AB sin ADB NOW, BD sin BAD sin {180° - (8 + x)}" sin

sing

11). a = sin ( + xj.. AB = a.sin (8 + x)**...(1). Again, ae = sin ABC = sin' a, .. AC = AB. sin as or from (1), AC = a. sin a.sin y D

sin (8 + x) 47. To find the distance of an object by observation from the top of a tower whose height is known.

Let B be the object in the same B horizontal plane with c the foot of the tower, and let the angle of depression DAB be observed.

Let AC = h, 4 DAB = 0.

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BC

Then we have, n = cot AR

= cot ABC = cot DAB, or
+ DARBC - osta

= cot e, :: BC = h cot 0, the distance required.

Cor. Suppose B to be not in the same level with C. Let m be the height of B above C, then B is on the same level with a point which is h m from A. . BC = (h m) cot e.

Ex. IX. 1. Find the height of a tower 200 yards distant when it subtends an angle of 15°.

2. From the top of a tower, the angle of depression of a point in the horizontal plane at the foot of the tower was 30°. Given the height to be 60 ft., find the distance of the point.

3. The angle of depression of two consecutive milestones in a direct line with the summit of a hill were observed to be 60° and 30°. Find the height of the hill.

4. An object is observed from a ship to be due E. After sailing due S. for six miles it is observed to be N.E. Find the distance from the last position of the ship.

5. There is an object A, and two stations B and C are taken in the same plane. Given that BC = 50, 2 ABC = 60°, 2 ACB = 30°, find AB, AC.

6. The elevation of a tower is found to be 45°, and on approaching 60 feet nearer the elevation is 75°. Find the height of the tower.

7. Wishing to know the breadth of a river I observed an object on the opposite bank, and, having walked along the side of the river a distance of 100 yards, found the angle subtended by the object and my first station to be 30°. Find the breadth of the river.

8. A person, standing exactly opposite to the centre of an oblong which measures 16 ft. by 12 ft., and such that the line drawn from the centre to his eye is at right angles to the oblong, observes that the diagonal subtends an angle of 60°. Find his distance.

9. The angles of elevation of the summit of one tower, whose height is h, are observed from the base and summit of another, and found to be o and the height of the second tower

respectively. Show that

= h.

tan o

"tano – tand 10. From the top of a tower the angles of depression of two objects in a direct line, and whose distance from each other is a, are a, ß respectively. Show that the height of the tower

a

S

cot B - cota

1

11. A person, having walked a distance a from one corner along a side of an oblong, observes that the side immediately behind him subtends an angle a, and the side in front an angle B. Show that the dimensions of the oblong are

a tan a, a (1 + tan a cot B). 12. Three points A, B, C form a triangle whose sides are a, b, c respectively, and a person standing at a point S, such that SA is at right angles to BC, observes that the side AC subtends an angle e. Show that the distance of S from B

= 2a Vla? + c – 62)2 + (a+ 62 c?) coť o. 13. A person walks a yards from A to E along AB the side of a triangle ABC, and observes that the angle AEC = a; he also walks b yards from B to F along BA, and observes Z CFB = B. Having given AB = c, find BC and AC.

14. A tower is observed from three stations A, B, C, in a straight line not meeting the tower, to subtend angles a, b, y respectively. Show that if AB = a, BC = b, the height of the tower

abla + b) Na cot”, – (a + b) cot* B + b cota When are the conditions impossible?

15. From two stations whose distance apart is a, and which are due W. and due S. respectively of one end of a wall, the angles subtended by the wall are each a. Show that the length of the wall is a sin a.

16. The angles of elevation of the top of a tower, whose height is h and standing on a hill, are a, b, when observed from two stations a miles distant, and in a direct line up the hill, Show that if o be the slope of the hill

a sin a sin B

cos 0 = Ti sin (- a)

17. The elevation of a tower was observed to be «, but on walking in the horizontal plane a distance a at right angles to the line joining the first position and the foot of the tower, the elevation was ß. Show that the height of the tower was

a

cot? ß – cot i 18. The angles of depression of two objects in the same horizontal plane, as seen from the top of a tower, are 0 and a respectively, and the angle they subtend is a. Show that if h be the height of the tower, the distance between the objects

= h N cosec 0 + cosec 0 - 2 cosec 0 cosec ° COS a.

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