247Question 1: Find AD :

Answer:

\frac{AE}{BE} = \tan 32^o

\Rightarrow AE = 20 \times \tan 32^o = 12.

AD = 5 + 12.5 = 17.5 \ m

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246Question 2: In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 \ m and \angle B = 60^o . Calculate the length of the board AB .

Answer:

\frac{PS}{PB} = \sin 60^o

PB = \frac{1}{\sin 60^o} = 1.15

\frac{PQ}{AP} = \cos 60^o

AP = \frac{1}{\cos 60^o} = 2

\therefore AB = PB + AP = 1.15 + 3 = 3.15 \ m

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245Question 3: Calculate BC .

Answer:

\frac{CD}{20} = \tan 42^o

CD = 20 \tan 42^o = 18

\frac{20}{BC+18} = \tan 35^o

20 = BC \tan 35^o + 18 \tan 35^o

BC = \frac{20 - 18 \tan 35^o}{\tan 35^o} = 10.56

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244Question 4: Calculate AB .

Answer:

\frac{AX}{6} = \cos 30^o

AX = 6 \cos 30^o = 5.196 \ m

\frac{XB}{5} = \cos 43^o

XB = 5 \cos 43^o = 3.856 \ m

AB = 5.196 + 3.856 = 8.85 \ m

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Question 5: The radius of a circle is given as 15 \ cm and chord AB subtends an angle of 131^o at the center C of the circle. Using trigonometry, calculate: (i) the length of AB ; (ii) the distance of AB from the center C .

Answer:

\frac{AX}{15} = \cos 24.5^o

AX = 15 \cos 24.5^o = 13.64 \ cm

Therefore AB = 2 \times 13.64 = 27.3 \ cm

\frac{OX}{15} = \sin 24.5^o

OX = 15 \sin 24.5^o = 6.22 \ cm

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Question 6: At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \frac{5}{12} on walking 192 meters towards the tower; the tangent of the angle is found to be \frac{3}{4} Find the height of the tower.

Answer:

\frac{h}{DB} = \frac{3}{4} \Rightarrow h = DB . \frac{3}{4}

\frac{h}{CB} = \frac{5}{12} \Rightarrow h = \frac{5}{12} (192+DB)

Therefore

\frac{5}{12} (192+DB) = DB . \frac{3}{4}

960 + 5 DB = 9 DB \Rightarrow DB = 240 \ m

Hence h = 240 \times \frac{3}{4} = 180 \ m

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Question 7: A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h meter. At a point on the plane, the angle of elevation of the bottom of the flagstaff is \alpha and that of the top of flagstaff is \beta . Prove that the height of the tower is \frac{h \tan \alpha}{\tan \beta - \tan \alpha} .

Answer:2019-05-20_6-46-25

\frac{h+DB}{CB} = \tan \beta

\frac{DB}{CB} = \tan \alpha

CB = \frac{DB}{\tan \alpha}

\frac{h + DB}{\frac{DB}{\tan \alpha}} = \tan \beta

h + DB = DB \Big( \frac{ \tan \beta}{\tan \alpha} \Big)

\Rightarrow h = \Big( \frac{\tan \beta - \tan \alpha}{\tan \alpha} \Big) DB

\therefore DB = \Big( \frac{\tan \alpha}{\tan \beta - \tan \alpha} \Big) h

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24x.jpgQuestion 8: With reference to the given figure of a man stands on the ground at point A , which is on the same horizontal plane as B , the foot of the vertical pole BC . The height of the pole is 10 \ m . The man’s eye is 2 \ m above the ground. He observes the angle of elevation of C , the top of the pole, as x^o where \tan x = \frac{2}{5} Calculate: (i) the distance AB in meters; (ii) angle of elevation of the top of the pole when he is standing 15 meters from the pole. Give your answer to the nearest degree.    [1999]

Answer:2019-05-20_6-52-36

\tan x = \frac{2}{5}

\frac{CE}{DE} = \frac{2}{5}

DE = \frac{5}{2} \times ( 10 - 2) = 20 m

i) AB = DE = 20 m

ii) if the distance from pole = 15 m

\therefore \tan y = \frac{8}{15} = 0.5333 \Rightarrow y = 28.072^o

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Question 9: The angles of elevation of the top of a tower from two points on the ground at distances a and b meters from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is \sqrt{ab} meter.

Answer:2019-05-20_6-55-29

\tan \alpha = \frac{h}{a}

\tan \beta = \frac{h}{b} 

\alpha + \beta = 90^o

\tan (\alpha + \beta) = \tan (90^o)

\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha . \tan \beta} = \tan (90^o)

\Rightarrow 1 - \tan \alpha . \tan \beta = 0

\frac{h}{a} . \frac{h}{b} = 1

\Rightarrow h^2 = ab

\Rightarrow h = \sqrt{ab}

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242Question 10: From a window A , 10 \ m above the ground the angle of elevation of the top C of a tower is x^o , where \tan x^o = \frac{5}{2} and the angle of depression of the foot D of the tower is y^o where \tan y^o = \frac{1}{4} . See the given figure. Calculate the height CD of the tower in meters.    [2000]

Answer:

\tan x  = \frac{5}{2} \Rightarrow \frac{EC}{AE} = \frac{5}{2}

\tan y = \frac{1}{4} \Rightarrow \frac{ED}{AE} = \frac{1}{4} \Rightarrow AE = 4 \times 10 = 10\ m

\therefore EC = 40 \times \frac{5}{2} = 100 \ m

\therefore CD = ED + EC = 10 + 100 = 110 \ m

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Question 11: A vertical tower is 20 \ m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53 . How far is he standing from the foot of the tower?

Answer:2019-05-19_20-05-25

\cos \theta = 0.53

\Rightarrow \frac{AB}{CB} 

CB = \sqrt{20^2 + x^2}

\therefore x = 0.53 \sqrt{20^2 + x^2}

3.56 x^2 = 20^2 + x^2

\Rightarrow x = \sqrt{ \frac{400}{2.56} } 

\Rightarrow x = 12.5 \ m

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Question 12: A man standing on the bank of a river observes that the angle of elevation of the top of the tower is 60^o . When he moves 50 \ m away from the bank. he finds the angle of elevation to be 30^o . Calculate: (i) the width of the river and (ii) the height of the tree.    [2003]

Answer:2019-05-19_19-58-21

Let the height of the tree = h

Let the width of the river = w

\therefore \frac{h}{w} = \tan 60^o = \sqrt{3}

\frac{h}{w+50} = \tan 30^o = \frac{1}{\sqrt{3}}

\therefore h = \sqrt{3} w

\Rightarrow \sqrt{3} h = w + 50

\Rightarrow \sqrt{3} (\sqrt{3} w) = w + 50

\Rightarrow 3w = w + 50

\Rightarrow 2w = 50

\Rightarrow w = 25 \ m

\therefore h = 25\sqrt{3} = 43.3 \ m

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Question 13: A 20 \ m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is 60^o and angle of elevation of the top of the pole as seen from the foot of the tower is 30^o . Find: (i) the height of the tower (ii) the horizontal distance between the pole and the tower.

Answer:2019-05-19_19-54-54

\frac{20}{CB} = \tan 30^o = \frac{1}{\sqrt{3}}

CB = 20\sqrt{3} \ m

\therefore \frac{AB}{CB} = \tan 60^o = \sqrt{3}

\Rightarrow AB = 20 \sqrt{3} \times \sqrt{3} = 60 \ m

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Question 14: A vertical pole and vertical tower are on the same ground level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60^o and the angle of depression of the bottom of the tower is 30^o Find: (i) the height of the tower, if the height of the pole is 20 \ m ;(ii)the height of the pole; if the height of the tower is 75 \ m .

Answer:

i) \frac{DE}{AE} = \tan 60 \Rightarrow DE = AE (\sqrt{3}) 2019-05-20_8-23-03

\frac{EC}{AE} = \tan 30 \Rightarrow 20 \sqrt{3}

\therefore DE = 60 m

Height of tower = 60 + 20 = 80 m

ii) \frac{75-h}{AE} = \tan 60 \Rightarrow 75 - h = \sqrt{3} AE 2019-05-20_8-23-28

\frac{h}{AE} = \tan 30 \Rightarrow AE = \sqrt{3} h

\therefore 75 - h = \sqrt{3} (\sqrt{3} h)

75 = 4 - h

h = 18.5 m

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Question 15: From a point, 36 \ m above the surface of a lake, the angle of elevation of a bird is observed to be 30^o and angle of depression of its image in the water of the lake is observed to be 60^o . Find the actual height of the bird above the surface of the lake.

Answer:2019-05-20_8-28-20

\frac{h}{BC} = \tan 30 \Rightarrow BC = \sqrt{3} h

\frac{EF + 36}{BC} = \tan 60

\Rightarrow EF + 36 = \sqrt{3} h ( \sqrt{3}) = 3h

\therefore EF = 3h - 36

Now, AE = EF

\Rightarrow h + 36 = 3h - 36

\Rightarrow 72 = 2h

\Rightarrow h = 36 m

Therefore height of the bird = 72 m

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Question 16: A man observes the angle of elevation of the top of a building to be 30^o . He walks towards it in a horizontal line through its base. On covering 60 \ m , the angle of elevation changes to 60^o . Find the height of the building correct to the nearest meter.

Answer:2019-05-20_8-29-56

\frac{h}{CB} = \tan 60 \Rightarrow CB = \frac{h}{\sqrt{3}}

\frac{h}{DB} = \tan 30 \Rightarrow DB = \sqrt{3} h

\therefore \sqrt{3} h - \frac{h}{\sqrt{3}} = 60

\Rightarrow h \Big( \frac{3-1}{\sqrt{3}} \Big) = 60

h = 30 \sqrt{3} = 51.96 m \approx 52 m

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Question 17: As observed from the top of a, 80 \ m tall lighthouse, the angles of depression of two ships, on the same side of the light house in horizontal line with its base, are 30^o and 40^o respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.    [2012]

Answer:2019-05-20_8-34-52

\frac{80}{BC} = \tan 40 \Rightarrow \frac{80}{0.839} = 95.34

\frac{80}{BC+CD} = \tan 30

\Rightarrow 80\sqrt{3} = 95.34 + CD

CD = 43.22 m \approx 43 m

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Question 18: In the given figure, from the top of a building AB = 60 \ m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30^o and 60^o respectively. Find: (i) the horizontal distance between AB and CD . (ii) the height of the lamp Post.2019-05-20_8-40-32

Answer:

i) \frac{60}{BC} = \tan 60 \Rightarrow BC = \frac{60}{\sqrt{3}} = 20 \sqrt{3} = 34.34 m

ii) \frac{60-h}{20\sqrt{3}} = \tan 30 \Rightarrow \frac{60-h}{20\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow h = 40 m

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Question 19: An airplane, at an altitude of 250 \ m observes the angles of depression of two boats on the opposite banks of a river to be 45^o and 60^o respectively. Find the width of the river. Write the answer correct to the nearest whole number.    [2014]

Answer:2019-05-20_8-47-27

\frac{250}{BD} = \tan 45 \Rightarrow BD = 250 m

\frac{250}{CD} = \tan 60 \Rightarrow CD = \frac{250}{\sqrt{3}} 

Therefore the width of the river

= BD + CD = 250 + \frac{250}{\sqrt{3}} = 250 + 144.34 = 394.34 m