trigo 1 class test {ms}

8/20/2019 Trigo 1 Class Test {MS}

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1. (a) area of AOP = 2

1

r 2 sin θ A1

(b) TP = r tan θ (M1)

area of POT = 21

r (r t an θ )

= 2

1

r 2 t an θ A1

(c) area of sector OAP = 2

1

r 2 θ A1

area of triangle OAP < area of sector OAP < area of triangle POT R1

θ θ θ tan

2

1

2

1sin

2

1 222r r r <<

sin θ < θ < tan θ AG[5]

IB Questionbank Mathematics Higher Level 3rd edition 1



2.

α = 2 arcsin

⇒

(

7

5.4

α = 1.396... = !.!1!° ...) M1(A1)

β = 2 arcsin

⇒

(5

5.4

β = 2.239... = 12.31°...) (A1)

Note: Allo" #se of cosine r#le.

area P = 2

1

× 72 × (α – sin α) = 1!.!... M1(A1)

area Q = 2

1

× 52 × ( β – sin β ) = 1.1... (A1)

Note: T$e M1 is for an atte%&t at area of sector %in#s area of triangle.Note: T$e #se of 'egrees correctl conerte' is acce&table.

area = 2.3(c%2) A1

[7]

3. A = 2

θ

(R 2 – r

2) A1

B =

2

2 r

θ

A1

fro% A: B = 2*1+ "e $ae R2 – r

2 = 2r

2M1

R = r 3 (A1)

$ence e,act al#e of t$e ratio R : r is 3 *1 A1 -![5]

IB Questionbank Mathematics Higher Level 3rd edition



4. A = A/ = 1! (c%) A1triangle O/ is e0#ilateral (M1)/ = 6 (c%) A1

EITHER

103arcsin2CA =

M1A1

CA = 3.9° (acce!t 0."0# ra$ians) A1

OR

200

1"4

10102

"1010CAcos

222

=××−+

=M1A1

CA = 3.9° (acce!t 0."0# ra$ians) A1

Note: Ot$er ali' %et$o's %a be seen.[6]

5. let t$e lengt$ of one si'e of t$e triangle be ! consi'er t$e triangle consisting of a si'e of t$e triangle an' t"o ra'ii

EITHER

! 2 = r

2 " r

2 – 2r

2 cos 12!° M1

= 3r 2

OR

! = 2r cos 3!° M1

THEN

! = 3r A1

so &eri%eter = 33 r A1no" consi'er t$e area of t$e triangle

area = 3 × 2

1

r 2sin 12!° M1

= 3 ×

2

43 r

A1

2

4

33

33

r

r

A

P =

= r

4

A1

Note: Acce&t alternatie %et$o's[6]

IB Questionbank Mathematics Higher Level 3rd edition 3



6. #se of cosine r#le* / =)70cos%727%( 22 ××−+

= .626... (M1)A1

Note: Acce&t an e,&ression for /2.

/ = .4614... ( = 2.!...) A1

#se of sine r#le*

= C

70sin7arcsinˆ B

= 9.61...° (C ˆ = 6!.34...°)(M1)A1

&se of cosine r&'e A = B%cos)2)%22 ××−+ = 6.12 (c%) A1

Note: 5cale 'ra"ing %et$o' not acce&table.[6]

7. (a) area = 2

1

× C × A × sin B (M1)

×××= Bsin"5

2110

3

2ˆsin = B

A1

(b) cos B = * 3

5

(= *0.7453...) or B = 1.... an' 13.1... (A1)

A2 = /

2 A/

2 – 2 × C × A × cos B (M1)

A = ...7453.0"52"5or...7453.0"52"52222 ×××++×××−+

A = .!3 or 1!.2 A1A1[6]

8. (a) = A 7 A = b 7 c cos A R1AG

(b) METHOD 1

/2 = /

2

2(M1)

a2 = (c sin A)2 (b 7 c cos A)2 (A1)

= c 2 sin

2 A b

2 7 2bc cos A c

2 cos

2 A A1

= b2 c

2 7 2bc cos A A1

METHOD 2

/2 = A/

2 7 A

2 = /

2 7

2(M1)(A1)

⇒ c 2 7 c

2 cos

2 A = a

2 7 b

2 2bc cos A 7 c

2 cos

2 A A1

⇒ a2 = b

2 c

2 7 2bc cos A A1

(c) METHOD 1

IB Questionbank Mathematics Higher Level 3rd edition #



b2 = a

2 c

2 7 2ac cos 6!° ⇒ b

2 = a

2 c

2 7 ac (M1)A1

⇒ c 2 7 ac a

2 7 b

2 = ! M1

⇒ c =

( ) ( )2

4 222baaa −−−±

(M1)A1

= 4

34

22

34 2222abaaba −

±=−±

(M1)A1

=

22

4

3

2

1aba −±

AG

Note: an'i'ates can onl obtain a %a,i%#% of t$e firstt$ree %ar8s if t$e verify t$at t$e ans"er gien int$e 0#estion satisfies t$e e0#ation.

METHOD 2

b2 = a

2 c

2 7 2ac cos 6!° ⇒ b

2 = a

2 c

2 7 ac (M1)A1

c 2 7 ac = b

2 7 a

2(M1)

c 2 7 ac

2

22

2

22

+−=

a

aba

M1A1

22

2

4

3

2ab

ac −=

−

(A1)

22

4

3

2ab

ac −±=−

A1

22

4

3

2

1abac −±=⇒

AG[12]

9. PR = h tan °+ R = h tan !° "$ere R5 = h M1A1A1

:se t$e cosine r#le in triangle PR. (M1)

2!2 = h

2 tan

2 ° h

2 tan

2 !° 7 2h tan ° h tan !° cos ° A1

°°°−°+°=

45cos50tan55tan250tan55tan

40022

2h

(A1)

= 349.9... (A1)

h = 19. (%) A1[8]

IB Questionbank Mathematics Higher Level 3rd edition $

trigo 1 class test {ms}

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