Question 1: 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

\\

242Question 2: 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

\\

Question 3: 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

\\

Question 4: 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

\\

Question 5: 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

\\

Question 6: From the top of a light house 100 \ m high, the angles of depression of two ships are observed as 48^o and 36^o respectively. Find the distance between the two ships (in the nearest meter) if:

(i) the ships are on the same side of the light house,

(ii) the ships are on the opposite sides of the light house. [2010]

33Answer:

(i) From \triangle ABC

\frac{100}{BC} = tan \ 48^o \Rightarrow BC = \frac{100}{tan \ 48^o}

Similarly, from \triangle ADB

\frac{100}{DB} = tan \ 36^o \Rightarrow DB = \frac{100}{tan \ 36^o}

Therefore, the distance between the ships is =

DB - BC  = \frac{100}{tan \ 36^o} -  \frac{100}{tan \ 48^o}

= 100 ( \frac{tan \ 48^o - tan \ 36^o}{ tan \ 48^o . tan \ 36^o})

= \frac{100 \times 0.3841}{0.8069} = 47.60 \approx 48  \ m

(ii) If the ships were on the opposite sides, then the distance between the ships is =

DB + BC = \frac{100}{tan \ 36^o} +  \frac{100}{tan \ 48^o}

= 90.040 + 136.63 = 227.68 \approx 228 \ m

\\

Question 7: From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be 30^o and 45^o respectively. Find the distances of the two stones from the foot of the hill. [2007]

314Answer:

From \triangle ABC

\frac{h}{BC} = tan \ 45^o \Rightarrow h = BC 

Similarly, from \triangle ABD

\frac{h}{1 + BC} = tan \ 30^o

BC = (BC + 1) \ tan \ 30^o 

BC \ ( 1 - tan \ 30^o) = tan \ 30^o 

BC = \frac{tan \ 30^o}{1 - tan \ 30^o} = \frac{1}{\sqrt{3} - 1} = 1.366 \ km 

Therefore BC = 1.366 \ km \ and \  DC = 2.366 \ km