Trigonometry

Trigonometry

(A) Problems on Triangles 1. A triangle ABC has the sides a, b, c according to A = 40° , a = 6.7cm, c = 8.5cm, the usual notation. If find the possible values for the remaining angles and side of the triangle Solutio n:

C a= 6.7cm

400 A c= 8.5cm

B

a c = sin A sin C 6.7 8.5 = sin 40° sin C sin C = 0.8155 ∠C = 54.6°

∠B = 180° − 40° − 54.6° = 85.4°

a b = sin A sin B 6.7 b = → b = 10.39cm sin 40° sin 85.4° ∴ ∠B = 85.4°, ∠C = 54.6° and b = 10.39cm

2. In the triangle ABC, AB = 5cm, BC = 10cm, CA = 8cm, If X is the mid-point of the median AA’ , find the length of BX. Solution A' B = A' C = 5cm : C X A

5cm

m 10c

8cm

AC 2 = AB 2 + BC 2 − 2( AB)( BC ) cos B

B

82 = 52 + 102 − 2(5)(10) cos B B = 52.4°

A ’

AA'2 = AB 2 + A' B 2 − 2( AB)( A' B ) cos B AA'2 = 52 + 52 − 2(5)(5) cos 52.4° = 19.4927 AA' = 4.415cm 1 A' X = AA' = 2.2075cm 2

1 A' = (180 − 52.4°) = 63.8° 2 BX 2 = A' B 2 + A' X 2 − 2( A' B)( A' X ) cos A' BX 2 = 52 + 2.20752 − 2(5)(2.2075) cos 63.8° = 20.1268 BX = 4.49cm

3. a

D

C

P b

b

A

a

B

For rectangle ABCD shown in the figure above, AB = BC = b. The foot of the perpendicular from A to BD is P. Calculate the lengths of AP and CP.

Solution :

BD = AD 2 + AB 2 = a 2 + b 2 AD b sin ∠ABD = = BD a 2 + b2 AP AP sin ∠ABD = = AB a b AP ab = → AP = a a 2 + b2 a 2 + b2

AD 2 = AP 2 + PD 2  ab b =  2 2  a +b 2

(

2

  + PD 2 

)

b 2 a 2 + b 2 − a 2b 2 b4 PD = = 2 2 2 a +b a + b2 2

b4 b2 PD = = 2 2 a +b a 2 + b2

a

D

C

P

∠CDP = ∠ABD then cos ∠CDP =

a a 2 + b2

CP 2 = CD 2 + DP 2 − 2(CD )( DP) cos ∠CDP

CP 2 = CD 2 + DP 2 − 2(CD )( DP) cos ∠CDP 2

 b 2 = a +  2 2  a +b

  b2  − 2a 2 2   a +b

2

(

)

 a  2 2  a + b

a 2 a 2 + b 2 + b 4 − 2 a 2b 2 = a 2 + b2

(

)

a 4 + a 2b 2 + b 4 − 2 a 2b 2 a 4 + b 4 − a 2 b 2 = = 2 2 a +b a 2 + b2

(

)

a +b −a b   CP =  2 2 a +b   4

4

(

2 2

)

(

1

2

)

  

STPM’ 96

4. In the triangle PQR, PQ = a, QR = a + d, RP ∠PQR > 120° = 2 , where a and a+2d, and a < d < a. d are 3 positive constants. Show that Solution : P a+ 2d

a

>1200 Q

a+d

R

I PR 2 > PQ 2 + QR 2 − 2( PQ)(QR ) cos ∠PQR

( a + 2d ) 2 > (a)2 + (a + d )2 − 2(a)(a + d ) cos120°

(

) (

)

 1 a 2 + 4ad + d 2 > a 2 + a 2 + 2ad + d 2 − 2 a 2 + ad  −   2 2a 2 − ad + 3d 2 < 0 (2a − 3d )(a + d ) < 0 3 −d
II PR 2 < PQ 2 + QR 2 − 2( PQ)(QR) cos180°

( a + 2d ) 2 < (a)2 + (a + d )2 − 2(a)(a + d ) cos180° a 2 + 4ad + 4d 2 rel="nofollow"> a 2 + ( a 2 + 2ad + d 2 ) − 2( a 2 + ad )( − 1) 3d 2 < 3a 2 d
I

+ II

2 a
STPM’ 97

5. In the triangle ABC, AB = p, BC = 3p, and ABˆ C = 60°. The bisector of the angle ABC meets AC at the point T. Find AT and CT in terms of p. Solution AC 2 = AB 2 + BC 2 − 2( AB)( BC ) cos B : A = p 2 + (3 p ) 2 − 2 p (3 p ) cos 60°

T

p

B

2

= 10 p − 6 p   2

60° 3p

2 1 

C

= 7 p2 AC = 7 p

AB AC = sin C sin B p 7p = sin C sin 60°  3 1 1 3  sin C =  =  2  7 2 7 −1  1

3   = 19.1066° C = sin  2 7

BC AC = sin A sin B 3p 7p = sin A sin 60°  3 3 3 3  sin A =  =  2  7 2 7 −1  3

3   = 100.8934° A = sin  2 7

3 3

2 7 A -1

cos A = −

1 2 7

Because of A = 180° − 60° − 19.1067° = 100.8934°

T = 180° − 30° − ∠A

sin T = sin (150° − ∠A)

= sin150°cos( ∠A) − cos150° sin ( ∠A) 1 1   3  3 3     = −  − −   2  2 7   2  2 7  1 9 2 =− + = 4 7 4 7 7

AT AB = sin 30° sin T AT p = ⇒ 1 2 2 7

AT =

CT = AC − AT = 7 − =

3 7 4

7 4

7 4

(B) Three Dimensional Problems 6. OAB is a triangle with angle AOB = 1200, OA = a , OB = 3a and ON which is perpendicular to AB, meets AB at N. By 3a considering the area of 3 = triangle OAB,ON or otherwise, show that

2 13

The line OP which is not on the plane AOB, is

PA2 = AN 2 + PN 2

such that OP is perpendicular with OA and OB, and OP = 2a. Prove that

Solution : O 120

a

0

A

N

3a

1 Area ∆OAB = (OA)(OB) sin ∠AOB 2 1 B = ( a )( 3a ) sin 120° 2 3 2 3   = a  2  2  3 3 2 = a 4

-----Eq(1)

AB 2 = OA2 + OB 2 − 2(OA)(OB ) cos ∠AOB = a 2 + (3a ) 2 − 2a (3a ) cos120° 1 = 10a − 6a  −   2 2

= 13a 2 AB = 13a

2

1 Area ∆OAB = (ON )( AB) 2 1 = ( ON ) 13a 2  13  =  a ( ON )  2 

(

) -----Eq(2)

Let Eq(1) = Eq(2)

3 3 2  13  a =  a ( ON ) 4  2  3a 3 ON = 2 13

(SHOWN)

P 2a O

a A

3a B

N

AP = OA +OP = a + ( 2a ) = 5a 2 2

2

2

2

2

2

 3 3  235 2 PN = OP + ON = ( 2a ) +  a  = a 52  2 13  2

2

2

2

27 2 25 2 AN = OA − ON = a − a = a 52 52 2

2

2

2

PN 2 + AN 2 =

235 2 25 2 a + a = 5a 2 = PA2 52 52

(proved)

Area ∆APB = s ( s − a )( s − b)( s − c)  1 5 a s = ( 13a + 13a + 5a ) =  13 + 2 2    5  5  5  5  a ×  a ×  a ×  13 − Area ∆APB =  13 + a  2   2   2   2   2 5  5  4 235 a  = 13 -  a = a 2 = 235 4  4  16 4 

7. A tetrahedron ABCD has four surfaces, each being an equilateral triangle with side of length 4a. If the mid-point of AB is L, find the angle between DL and DC. If X is a point on BC at a distance of a from C, calculate the lengths if XL and XD, and verify that 2 2 2 2

XL + LD + DX = 32a

If the perpendicular from D to LX at P, find the ratio LP : PX.

Solution :

A

L

Angle between DL and DC =∠LDC L 4a

B

D 4a

4a C

α C

2a

2a

D

=α

C

LC 2 + BL2 = BC 2 LC + ( 2a ) = ( 4a ) 2

4a

2

LC 2 = 12a 2 B

2a

L

2a

A

LC = 12a = 2 3a

2

2a 1 cos α = = 2 3a 3 −1  1  α = cos   = 54.7°  3

C

a X

4a

B

2a

L

2a

∠A = ∠B = ∠C = 60°

A

XL2 = LB 2 + BX 2 − 2( LB )( BX ) cos ∠B = ( 2a ) + ( 3a ) − 2( 2a )( 3a ) cos 60° 2

= 7a 2 XL = 7a

2

D

4a

B

C

X a

XD 2 = CX 2 + CD 2 − 2(CX )(CD ) cos ∠C = ( a ) + ( 4a ) − 2( a )( 4a ) cos 60° 2

= 13a 2 XD = 13a

2

Show

XL2 + LD 2 + DX 2 = 32a 2

XL2 + LD 2 + DX 2 = 7 a 2 + LC 2 + 13a 2 = 20a 2 + 12a 2 = 32a 2 (shown)

D

2 3a

13a

X P

L

7a

XD 2 = LX 2 + LD 2 − 2( LX )( LD ) cos ∠L

(

) ( 2

13a =

) ( 2

)

2

(

)(

)

7a + 2 3a − 2 7a 2 3a cos ∠L

7 a 2 + 12a 2 − 13a 2 3 cos ∠L = = 2 4 21a 2 21

LP cos ∠L = DL 3 LP = 2 21 2 3a 3 3  7 3 7  3   a = LP =  a= a 2 3a = 7 7 7 7  2 21  3 4 PX = LX − LP = 7a − 7a = 7a 7 7 LP : PX →

3 4 7a : 7a → 3 : 4 7 7

8. A circular cylinder has length l, and radius a, and is placed so that its axis is horizontal. The vertical diameter of one of its end is AB, with A above B. At the other end is the diameter CD. The angle between CD and the vertical is α; C lies above the level of D. Find the lengths of AC and AD. If

9 cos α = 16

l 2 25 = 2 8 , find the angles of the and a

triangle ACD.

A

Solution :

l

A

a

C

α

l C D

α O

B O a A ’

α

A ’

A' C 2 = a 2 + a 2 − 2( a )( a ) cosα = 2a 2 − 2a 2 cos α

a C

= 2a 2 (1 − cosα )

(

AC 2 = AA'2 + A' C 2 = l 2 + 2a 2 (1 − cosα )

)

α  2α 2α 2 α  = l + 2a  sin + cos −  cos − sin   2 2  2 2   α α  = l 2 + 2a 2  2 sin 2  = l 2 + 4a 2 sin 2 2 2  2

2

2

 2 2 2α  AC =  l + 4a sin  2 

1

2

A' D 2 = a 2 + a 2 − 2( a )( a ) cos(180° − α ) A

= 2a 2 − 2a 2 ( − cosα ) = 2a 2 (1 + cosα )

l C α D

A ’

α α α α  = 2a 2  sin 2 + cos 2 + cos 2 − sin 2  2 2 2 2  2 2α  = 4a  cos  2  α  AD 2 = AA'2 + A' D 2 = l 2 +  4a 2 cos 2  2  α  AD =  l 2 + 4a 2 cos 2  2 

1

2

α 9 −1 9 Given that cos α = α = cos = 55.77° → = 27.89° 16 2 16

α cos = cos 27.89° = 0.8839 2 α sin = sin 27.89° = 0.4678 2 and

l 2 25 = 2 8 a



25 2 l = a 8 2

1

 25 2 2 AC =  a + 4a 2 ( 0.4678)   8 

 25 2 2 AD =  a + 4a 2 ( 0.8839)   8 

2

1

2

( )

= 4a

2

1

2

(

= 6.25a

2

= 2a

)

1

2

= 2.5a

C

CD = 2a 2a

2a

A

D

2.5a

AD 2 = AC 2 + CD 2 − 2( AC )( CD ) cos ∠ACD

( 2.5a ) 2 = ( 2a ) 2 + ( 2a ) 2 − 2( 2a )( 2a ) cos ∠ACD 6.25a 2 − 4a 2 − 4a 2 cos ∠ACD = = 0.21875 2 − 8a ∠ACD = cos−1 0.21875 = 77.36° 180° − 77.36° ∠CAD = ∠ADC = = 51.32° 2

9. A cube is placed on a horizontal plane so that its surface ABCD touches the plane. Each side of the cube has a length of 4a. A piece of string, with one end attached to A, is tied around the four vertical surfaces of the cube and touches the vertical edges passing through B, C and D at the points L, M and N respectively. The other end of the string is tied to the vertex A’ located

Find a) The length of the straight line AM, b) The angle ALM, c) Cosine of the angle MAN, d) Area of the triangle MAN.

Solution :

M

a)

2a

A’ N

M

L

4a

C

B A

4a

D C 4a

A

4a

D

C A

AC 2 = ( 4a ) + ( 4a ) = 32a 2 2

2

AM 2 = AC 2 + MC 2 = 32a 2 + ( 2a ) = 36a 2 2

AM = 6a

b)

L

17a

6a

A

4a

A

M

17a

( 6a ) 2 = 17a 2 + 17a 2 − 2(

)

17a cos ∠ALM

L

M

a

a

B 2

)(

36a 2 − 34a 2 2 cos∠ALM = =− 2 34 − 34a ∠ALM = 93.4°

AL = ( 4a ) + ( a ) = 17a 2

17a

2

L 2

4a

ML2 = ( 4a ) + ( a ) = 17a 2 2

2

M

c)

6a

A

17a 5a

(

N

)

17a = ( 6a ) + ( 5a ) − 2( 6a )( 5a ) cos ∠MAN 2

2

2

17a 2 − 61a 2 − 44 11 cos∠MAN = = = 2 − 60 15 − 60a N 5a A

4a

3a D

d) 15

∠MAN

2 26

2 26 sin ∠MAN = 15

11

1 Area ∆MAN = ( AM )( AN ) sin ∠MAN 2 1 2 26 = ( 6a )( 5a ) = 2 26a 2 2 15

STPM’ 83

10. An aircraft flying in a straight line is climbing at an angle φ with the horizon. The aircraft is observed from a point O on the horizontal ground. The plane passes a point A’ located vertically above the point A, east of O and, it then passes the point B’ located vertically above B on the ground, north of O, where OA = OB. When the plane is at A’ , the angle of elevation of the plane from O is β. Prove that 1

tan ϕ =

2

( tan β − tan α )

Subsequently, show that when the aircraft is vertically above the mid-point of AB , its angle of elevation γ from O is given by

1 ( tan α + tan β ) tan γ = 2

Solution : B’ P

ϕ B

β

α O

A’

A

OB 1 sin 45° = → AB = OB AB 2 OA 1 cos 45° = → AB = OA AB 2

B 450

450 O

A

B’

BB' tan β = → OB

1 AB tan β = BB ' 2

β B

1 AB 2

O

AA' tan α = → OA

A’

1 AB tan α = AA' 2 O

α

1 AB 2

A

B’

A' P = AB

ϕ P

A’

1 1 B' P = BB'− AA' = AB tan β − AB tan α 2 2 1 = AB( tan β − tan α ) 2

1 AB( tan β − tan α ) B' P 1 2 ( tan β − tan α ) tan ϕ = = = A' P AB 2

B’ C’

A’ B

β

C

α

A

O

γ

C’ Q C

ϕ

C 'Q tan ϕ = AC CC ' − AA ' A’ = AC A

1 AC = BC = AB 2

C’

B 450

C

γ C

450 A O

1 So OC = BC = AB 2

1 AB 2

O

1 CC ' = OC tan γ = AB tan γ 2

From

CC '− AA' tan ϕ = AC 1 1 AB tan γ − tan α 1 2 ( tan β − tan α ) = 2 1 2 AB 2 1 1 2 tan β − tan α = tan γ − tan α 2 2 2

1 1 2 tan γ = tan β − tan α + tan α 2 2 2 1 1 = tan β + tan α 2 2 1 ( tan α + tan β ) = 2

11. The equilateral triangle ABC has sides each 6 units long. The altitudes of the AN = at 2 N, 3 units triangle meet show that . This triangle is the base of a pyramid with its vertex V vertically above N , so that the line VN is perpendicular to the plane ABC. ∠VAN = 30° Given that VN=2 units, prove that The perpendicular from A to the edge VC 3 7 meets AR = units, 2 cos ARB that CV extended at R.∠Prove and find the value of

N

Solution :

1200 A

A

6u

nit s

s nit 6u

C 6 units

300

B

6 units

N

B

300

AN 6 = sin 30° sin 120° AN 6 = 1 3 2 2 6  3   = 2 3 units AN = 3 3

V

A

V

2 units

B N A

C

2

2 3 units

1 tan ∠VAN = = 2 3 3 ∠VAN = 30° (proved)

N

un its

V

4

nits

8

2

)

2

2

C

62 = 42 + 42 − 2( 4 )( 4 ) cos ∠AVC 36 − 32 4 1 cos∠AVC = = =− − 32 − 32 8

63 = 3 7 -1

2

VA = 4 units

R

A 6u

(

VA = AN + VN = 2 3 + ( 2 ) = 16 2

sin ∠AVC =

3 7 8

R

From triangle VRA 3 7 units 2

AR sin ∠AVC = 4 3 7 3 7 AR = ×4= unit 8 2 2

A 2

3 7 units 2

6 units

B

3 7  3 7   3 7  3 7  2  +   − 2   cos ∠ARB 6 =   2   2   2  2  63 9 36 − 2 = 2 = −1 cos∠ARB = 63 63 7 − − 2 2