ABOUT TRIGONOMETRY

TRIGONOMETRY IS DRIVED FROM GREEK WORDS TRIGONON = THREE ANGLES & MATRON = MEASURES.

TRIGONOMETRY IS THE BRANCH OF MATHS DEALS WITH TRIANGLES , PARTICULARLY TRIANGLES IN A PLANE WHERE ONE ANGLE IS 90 .

TRIGONOMETRY SPECIFICALLY DEALS WITH THE REALATIONSHIP BETWEEN THE SIDES AND THE ANGLES OF TRIANGLES .

TRIGONOMETRY IS MOST IMPORTANT APPLICATION IN MATHS WHICH IS USED IN OUR DAILY LIFE .

TRIGONOMETRIC RATIOS

IN A RIGHT ANGLED TRIANGLE : SIN =SIDE OPPSITE TO ANGLE HYPOTENUSE COS =SIDE ADJACENT TO ANGLE HYPOTENUSE TAN = SIDE OPPSITE TO ANGLE SIDE ADJACENT TO ANGLE COSEC = 1 SIN SEC = 1 COS COT= 1 TAN

THE VALUES OF TRIGONOMETRIC RATIOS

A 0 30 45 60 90SIN A 0 1/2 1/√2 √3/2 1COS A 1 √3/2 1/√2 1/2 0TAN A 0 1/√3 1 √3 N.DCOSEC A N.D 2 √2 2/√3 1SEC A 1 2/√3 √2 2 N.DCOT A N.D √3 1 1/√3 O

RELATIONSHIP BETWEEN T RATIOS SIN (90-A) = COS A , COS (90-A) = SIN A

TAN (90-A) = COT A , COT (90-A) = TAN A

A THEODOLITE

(SURVEYING INSTUMENT , WHICH IS BASED ON THE PRINCIPLES OF TRIGONOMETRY , IS USED FOR MEASURING ANGLES WITH A ROTATING TELESCOPE )

SURVEYORS HAVE USED TRIGONOMETRY FOR CENTURIES . ONE SUCH LARGE SURVEYING PROJECT OF THE NINETEENTH CENTURY WAS THE “GREAT TRIGONOMETIC SURVEY” OF BRITISH INDIA FOR WHICH THE TWO LARGEST –EVER THEDOLITES WERE BUILT . DURING THE SURVEY IN 1852 , THE HIGEST MOUNTAIN IN THE WORLD WAS DISCOVERED . FROM A DISTANC E OF OVER 160 KM , THE PEAK WAS OBSERVED FROM SIX DIFFERENT STATIONS . IN 1856 , THIS PEAK WAS NAMED AFTER SIR GEORGE EVEREST , WHO HAD COMMISSIONED AND FIRST USED THE GIANT THEDOLITES . THE THEODOLITES ARE NOW ON DISPLAY IN THE MUSEUM OF THE SURVEY OF INDIA IN DEHRADUN

HEIGHTS AND DISTANCES

A B

C

LINE OF SIGHT

ANGLE OF ELEVATION

THE LINE OF SIGHT IS THE LINE DRAWN FROM THE EDGE OF AN OBSERVER TO THE POINT IN THE OBJECT VIEWED BY THE OBSERVER . THE ANGLE OF ELEVATION OF THE POINT VIEWED IS THE ANGLE FORMED BY THE LINE OF SIGHT WITH THE HORIZONTAL WHEN THE POINT BEING VIEWED IS ABOVE THE HORIZONTAL LEVEL.

LINE OF SIGHT

HORIZONTAL LEVEL

ANGLE OF DEPRESSION

OBECT

THE ANGLE OF DEPREESION OF A POINT ON THE OBJECT BEING VIEWED IS THE FORMED BY THE LINE OF SIGHT WITH THE HORIZONTAL WHEN THE POINT IS BELOW THE HORIZONTAL LEVEL

PROBLEMS BASED ON APPLICATION OF TRIGONOMETRY

Q1 . FROM A POINT ON THE GROUND 40 M AWAY FROM THE FOOT OF A TOWER , THT ANGLE OF ELEVATION OF THE TOWER IS 30° . THE ANGLE OF ELEVATION OF THE TOP OF A WATER TANK IS 45°. FIND THE HIEGHT OF THE TOWER & THE DEPTH OF THE TANK .

SOL: IN ∆ABD , WE HAVE TAN 45°= BD/AB => 1= (H+H1) /40 => H+H1= 40 …………(1) IN ∆ABC , WE HAVE TAN 30°=BC/AB => 1/√3 = H / 40 => H =40/√3 m = 40√3/3 m = 23.1 m => H=23.1 m SUBSTITUTING THE VALUE OF H IN (1) , WE HAVE 23.1 + H1 = 40 => H1 = (40-23.1)m = 16.9 m HENCE . THE HEIGHT OF THE TOWER IS H= 23.1 m AND THE DEPTH OF THE

TANK IS H1 = 16.9 m .

3045

A B

C

DH1

H

40 m

X

Y

P

R

Q

x m

40 m60°

45°

Q2 . THE ANGLE OF ELEVATION OF THE TOP Q OF A TOWER PQ FROM A POINT X ON THE GROUND IS 60° . AT A POINT Y , 40m VERTICALLY ABOVE X , THE ANGLE OF ELEVATION IS 45°. FINDTHE HEIGHT OF THE TOWER PQ AND THE DISTANCE

SOL : IN ∆YRQ , WE HAVE TAN 45° = QR/YR => 1 = x / YR => YR = x => XP = x IN ∆XPQ , WE HAVE TAN 60° = PQ /PX => √3 = (x + 40)/x => √3x = x + 40 => x( √3 - 1 ) = 40 => x = (40/(√3-1))*((√3+1)/(√3+1)) = 20*(√3+1)= 54.64 SO , HEIGHT OF THE TOWER PQ = x + 40 = 54.64 + 40 = 94.64 m IN ∆XPQ , WE HAVE SIN 60° = PQ / XQ √3/2 = 94.64/ XQ ; XQ = (94.64*2)/√3 XQ = 109.3 m

30

30

45

45

Q3 . AS OBSERVED FROM THE TOP OF A LIGHT HOUSE , 100m ABOVE SEA LEVEL , THE ANGLE OF DPRESSION OF A SHIP , SAILING DIRECTLY TOWARDS IT , CHANGES FROM 30° & 45° . DETERMINE THE DISTANCE TRAVELLED BY THE SHIP DURING THE PERIOD OF OBERVATION .

100m

d A BP

O

SOL : LET A & B BE THE TWO POSITIONS OF THE SHIP . LET d BE THE DISTANCE TRAVELLED BY THE SHIP DURING THE PERIOD OF OBSERVATION i.e. AB =d mLET THE OBSERVER BE AT O , THE TOP OF THE LIGHT HOUSE PO .IT IS GIVEN THAT PO= 100 m & THE ANGLES OF DEPRESSION FROM O OF A AND B ARE 30° AND 45°RESP. OAP = 30° & OBP = 45° IN ∆OPB , WE HAVE TAN 45° =OP/BPÞ 1 = 100/BPÞ BP= 100 m IN ∆OPA , WE HAVEÞ TAN 30° =OP/APÞ 1/√3 = 100 / (d +BP)Þ d + BP = 100√3Þ d = 100√3 − 100Þ d = 100(√3 − 1 ) = 100(1.732−1) = 73.2 m HENCE , THE DISTANCE TRAVELLED BY THE SHIP FROM A TO B IS 73.2 m.

30

6060 m

h

60 m

h+60 A B

c`

c

MP

Q4 . THE ANGLE OF ELEVATION OF A CLOUD FROM A POINT 60 m ABOVE A LAKE IS 30° AND ANGLE OF DEPRESSION OF THE REFLECTION OF CLOUD IN THE LAKE IS 60°.FIND THE HEIGHT OF THE CLOUD .

SOL : LET PM BE THE POSITION OF THE LAKE AND P BE THE POINT OF OBSERVATION SUCH THAT AP = 60 m . LET C BE THE POSITION OF THE CLOUD AND C` BE ITS REFLECTION IN THE LAKE .LET CM = h THEN , CB = h+60 IN ∆CMP , WE HAVE TAN 30° = CM / PM => 1/√3 = h / PM => PM = √3h ………(1) IN ∆PMC` , WE HAVE TAN 60° = C`M/PM=> TAN 60° = (C`B+BM)/PM √3 = (h+60+60)/PM=> PM = (h+120)/√3 ………(2) FROM EQUATION (1) & (2) , WE GET √3h = (h+120)/√3=> 3h = h + 120 => 2h = 120 => h = 60 NOW , CB=CM + MB = h + 60 = 60+60 = 120HENCE , THE HEIGHT OF THE CLOUD FROM THE SURFACE OF THE LAKE IS 120 m.