HEIGHTS AND DISTANCES

IMPORTANT FACTS AND FORMULAE

hypotenuse

1.We already know that:

In a rt.angled DOAB, where ÐBOA = q,

i)sin q = Perpendicular/Hypotenuse = AB/OB;

ii)cos q = Base/Hypotenuse = OA/OB;

iii)tan q = Perpendicular/Base = AB/OA;

iv)cosec q = 1/sin q = OB/AB;

Base

v)sec q = 1/cos q = OB/OA;

vi)cot q = 1/tan q = OA/AB.

2. Trigonometrical identities:

i)sin²q + cos²q = 1.

ii)1+tan²q = sec²q

iii)1+cot²q = cosec²q

3. Values of T-ratios:-

q	0	30°	45°	60°	90°
Sin q	0	½	1/Ö2	Ö3/2	1
Cos q	1	Ö3/2	1/Ö2	½	0
Tan q	0	1/Ö3	1	Ö3	Not defined

4. Angle of Elevation: Suppose a man from a point O looks up an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

\ Angle of elevation of P from O = ÐAOP.

5. Angle of Depression: Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

SOLVED EXAMPLES

Ex.1.If the height of a pole is 2Ö3 metres and the length of its shadow is 2 metres, find the angle of elevation of the sun.

2Ö3m

C 2m A

Sol. Let AB be the pole and AC be its shadow.

Let angle of elevation, ÐACB=q.

Then, AB = 2Ö3 m AC = 2 m.

Tan q = AB/AC = 2Ö3/2 = Ö3 Þ q = 60°

So, the angle of elevation is 60°

Ex.2. A ladder leaning against a wall makes an angle of 60° with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall.

19m

60°

C A

Sol. Let AB be the wall and BC be the ladder.

Then, ÐACB = 60° and BC = 19 m.

Let AC = x metres

AC/BC = cos 60° Þ x/19 = ½ Þ x=19/2 = 9.5

\Distance of the foot of the ladder from the wall = 9.5 m

Ex.3. The angle of elevation of the top of a tower at a point on the ground is 30°. On walking 24 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

30° 60°

C 24m D A

Sol. Let AB be the tower and C and D be the points of observation. Then,

AB/AD = tan 60° = Ö3 => AD = AB/Ö3 = h/Ö3

AB/AC = tan 30° = 1/Ö3 AC=AB x Ö3 = hÖ3

CD = (AC-AD) = (hÖ3-h/Ö3)

hÖ3-h/Ö3 = 24 => h=12Ö3 = (12´1.73) = 20.76

Hence, the height of the tower is 20.76 m.

Ex.4. A man standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retires 36 m from the bank, he finds the angle to be 30°. Find the breadth of the river.

30° 60°

C 36m D x A

Sol. Let AB be the tree and AC be the river. Let C and D be the two positions of the man. Then,

ÐACB=60°, ÐADB=30° and CD=36 m.

Let AB=h metres and AC=x metres.

Then, AD=(36+x)metres.

AB/AD=tan 30°=1/Ö3 => h/(36+x)=1/Ö3

h=(36+x)/ Ö3 .....(1)

AB/AC=tan 60°=Ö3 => h/x=Ö3

h=Ö3x .....(2)

From (i) and (ii), we get:

(36+x)/ Ö3= Ö3x => x=18 m.

So, the breadth of the river = 18 m.

Ex.5. A man on the top of a tower, standing on the seashore finds that a boat coming towards him takes 10 minutes for the angle of depression to change from 30° to 60°. Find the time taken by the boat to reach the shore from this position.

30° 60°

D A

x C y

Sol. Let AB be the tower and C and D be the two positions of the boat.

Let AB=h, CD=x and AD=y.

h/y=tan 60°=Ö3 => y=h/Ö3

h/(x+y)=tan 30° = 1/Ö3 => x+y=Ö3h

x=(x+y)-y = (Ö3h-h/Ö3)=2h/Ö3

Now, 2h/Ö3is covered in 10 min.

h/Ö3will be covered in (10´(Ö3/2h)´(h/Ö3))=5 min

Hence, required time = 5 minutes.

Ex 6. There are two temples, one on each bank of a river, just opposite to each other. One temple is 54 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.

30°

D E

60°

C A

Sol. Let AB and CD be the two temples and AC be the river.

Then, AB = 54 m.

Let AC = x metres and CD=h metres.

ÐACB=60°, ÐEDB=30°

AB/AC=tan 60°=Ö3

AC=AB/Ö3=54/Ö3=(54/Ö3´Ö3/Ö3)=18m

DE=AC=18Ö3

BE/DE=tan 30°=1/Ö3

BE=(18Ö3´1/Ö3)=18 m

CD=AE=AB-BE=(54-18) m = 36 m.

So, Width of the river = AC = 18Ö3 m=18´1.73 m=31.14m

Height of the other temple = CD= 18 m.

Heights And Distances

probability

important facts and formula

1.Experiment :An operation which can produce some well-defined outcome is called an experiment

2.Random experiment: An experiment in which all possible outcome are known and the exact out put cannot be predicted in advance is called an random experiment

Eg of performing random experiment:

(i)rolling an unbiased dice

(ii)tossing a fair coin

(iii)drawing a card from a pack of well shuffled card

(iv)picking up a ball of certain color from a bag containing ball of different colors

Details:

(i)when we throw a coin. Then either a head(h) or a tail (t) appears.

(ii)a dice is a solid cube, having 6 faces ,marked 1,2,3,4,5,6 respectively when we throw a die , the outcome is the number that appear on its top face .

(iii)a pack of cards has 52 cards it has 13 cards of each suit ,namely spades, clubs ,hearts and diamonds

Cards of spades and clubs are black cards

Cards of hearts and diamonds are red cards

There are 4 honors of each suit

These are aces ,king ,queen and jack

These are called face cards

3.Sample space :When we perform an experiment ,then the set S of all possible outcome is called the sample space

eg of sample space:

(i)in tossing a coin ,s={h,t}

(ii)if two coin are tossed ,then s={hh,tt,ht,th}.

(iii)in rolling a die we have,s={1,2,3,4,5,6}.

4.event:Any subset of a sample space.

5.Probability of occurrence of an event.

let S be the sample space and E be the event .

then,EÍS.

P(E)=n(E)/n(S).

6.Results on Probability:

(i)P(S) = 1 (ii)0<P(E)<1 (iii)P(f)=0

(iv)For any event a and b, we have:

P(aÈb)=P(a)+P(b)-P(aÈb)

(v)If A denotes (not-a),then P(A)=1-P(A).

SOLVED EXAMPLES

Ex 1. In a throw of a coin ,find the probability of getting a head.

sol. Here s={h,t} and e={h}.

P(E)=n(E)/n(S)=1/2

Ex2.Two unbiased coin are tossed .what is the probability of getting atmost one head?

sol.Here s={hh,ht,th,tt}

Let Ee=event of getting one head

e={tt,ht,th}

p(e)=n(e)/n(s)=3/4

Ex3.An unbiased die is tossed .find the probability of getting a multiple of 3

sol. Here s={1,2,3,4,5,6}

Let e be the event of getting the multiple of 3

then ,e={3,6}

p(e)=n(e)/n(s)=2/6=1/3

ex4. in a simultaneous throw of pair of dice .find the probability of getting the total more than 7

sol. Here n(s)=(6*6)=36

let e=event of getting a total more than 7

={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

p(e)=n(e)/n(s)=15/36=5/12.

Ex5. A bag contains 6 white and 4 black balls .2 balls are drawn at random. find the probability that they are of same colour.

Sol .let s be the sample space

Then n(s)=no of ways of drawing 2 balls out of (6+4)=10c2=(10*9)/(2*1)=45

Let e=event of getting both balls of same colour

Then n(e)=no of ways(2 balls out of six) or(2 balls out of 4)

=(⁶c2+⁴c2)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21

p(e)=n(e)/n(s)=21/45=7/15

Ex6.Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6

sol. Clearly n(s)=6*6=36

Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then

e={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),

(6,6)}

n(e)=14.

Hence p(e)=n(e)/n(s)=14/36=7/18

Ex7.Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

sol. We have n(s)=52c2=(52*51)/(2*1)=1326.

Let A=event of getting both black cards

B=event of getting both queens

aÇb=event of getting queen of black cards

n(A)=²⁶c2=(26*25)/(2*1)=325,

n(b)=⁴c2=(4*3)/(2*1)=6 and

n(aÇb)=2c2=1

p(A)=n(A)/n(S)=325/1326;

p(B)=n(B)/n(S)=6/1326 and

p(aÇb)=n(aÇb)/n(s)=1/1326

p(aÈb)=p(a)+p(b)-p(aÇb)=(325+6-1/1326)=330/1326=55/221