Contoh Soal Luas Segitiga Dalam Trigonometri

Soal-soal Luas Segitiga dalam Trigonometri 1. a. Untuk Untuk mencar mencarii rumu rumus s b2 b

a

C

Cos

t

Β=

BD

Sin

BC

Β=

CD BC BC

sin Β BC ⋅ cos Β CD = BC ⋅ sin = BC BD = a ⋅ cos Β sin Β CD = a ⋅ sin BD

c

α

D

A

B

= CD 2 + AD 2 AD = AB − BD AD = c − ( a ⋅ cos Β ) 2 2 2 sin Β) + ( c − a ⋅ cos Β) AC = ( a ⋅ sin 2 2 sin 2 Β + ( c 2 − 2 ac ⋅ cos Β + a 2 ⋅ cos 2 Β) b = a ⋅ sin 2 2 2 sin 2 Β + cos 2 Β ) b = c − 2 ac ⋅ cos Β + a ( sin 2 2 2 b = c − 2 ac ⋅ cos Β + a 2 2 2 b = a + c − 2 ac ⋅ cos Β 2

AC

C

a

D t

c

b. Untuk mencari rumus c2

Sin C

=

BD BC

sin C = BC ⋅ sin sin C BD = a ⋅ sin BD

b

α Cos C

=

CD BC

= BC ⋅ cos C CD = a ⋅ cos C CD

A

B

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= BD 2 + AD 2 AD = AC − CD AD = b − ( a ⋅ cos C ) 2 2 2 sin C ) + ( b − a . cos C ) AB = ( a ⋅ sin 2 2 sin 2 C + ( b 2 − 2 ab cos C + a 2 ⋅ cos 2 C ) c = a ⋅ sin 2 2 2 sin 2 C + cos 2 C ) c = b − 2 ab ⋅ cos C + a COS ( sin 2 2 2 c = b − 2 ab ⋅ cos C + a 2 2 2 c = a + b − 2 ab ⋅ cos C AB

2

2. Bu Bukt ktik ikan an :

∆

L ABC

=

a

2

⋅ sin sin B ⋅ sin sin C 2 sin sin

A

a)

∆

L ABC

=

b

2

⋅ sin sin A ⋅ sin sin C 2 sin sin

B

b)

∆

L ABC

=

c

2

⋅ sin sin A ⋅ sin sin B 2 sin sin

C

c) Jawab : a

sin sin

A

=

Aturan Kosinus :

∆

L ABC

a)

=

a

2

⋅ sin sin B ⋅ sin sin 2 sin sin

A

C

b

sin sin

B

=

c

sin sin

C





b

Sin B

C

a

t

t

= a ⋅ sin sin B = 1 ac ⋅ sin sin B

∆

D

A

a

L ABC

α

c

t

=

∴

B

2

a

sin sin

=

A

=

1

∆

=

1

L ABC

∆

L ABC

∆

L ABC

=

2

⋅ sin sin A ⋅ sin sin 2 sin sin

sin sin

∆

L ABC

b

c C

sin sin 2

sin sin

C

2 sin sin

b) C b

a

t

=

t b

sin A = b ⋅ sin

∆

L ABC

t

α c

∴

D

A

b

sin sin

B

=

c

sin sin

∆

=

1

∆

=

1

∆

=

1

L ABC

L ABC

L ABC

C

bc

2

⇒c=

b

⋅ sin sin sin sin

B

 b ⋅ sin sin C   ⋅  sin  sin sin si n B   2 sin C ⋅ sin sin A b ⋅ sin

2 2

C

⋅ sin sin A

b

2

b

B

sin sin

sin sin

B

C

sin sin

A

A

=

1 2

A

bc

A

A

B

A

sin sin

⋅ sin sin B

C

sin sin

⋅ sin sin C

 a ⋅ sin   sin C  ⋅ sin sin B sin A   sin  2 sin C ⋅ sin sin B a ⋅ sin

2

=

a

a

2

a

⇒c=

⋅ sin sin A

(Terbukti





(Terbukti)

∆

L ABC

b

a

=

2

⋅ sin sin A ⋅ sin sin B

a 2 sin sin

c)

C

C

t

c

α D

A

B

Sin B t

=

∆

a

sin sin

A

∆

L ABC

=

=

c

sin sin

1 2

C

ac

⇒a=

c

sin A ⋅ sin sin sin

C

sin B ⋅ sin

1  c ⋅ sin sin A   ⋅ c ⋅ sin sin B =  2  sin sin C   2 sin A ⋅ sin sin B 1 c ⋅ sin L∆ ABC =

∆

L ABC

2

sin sin

C

a

= a ⋅ sin sin B

L AB ABC C

∴

t

=

1 2

ac

⋅ sin sin B





3. Cnt! Cnt! sa" sa"#s #sa" a" se$it se$it$a $a 40 3

1. Jika BC% 16 cm & AC%10 cm 'an "uas tentukan anan$ AB * Jawab :

∆ABC%

cm2& mak maka

C

∆

L ABC 16 cm

⇔ 40

10 cm

=

3

1

AC .CB

2 1

sin sin

C

sin C = ⋅ 10 ⋅ 16 sin

sin C = ⇔ sin

2

40 3 80

=

1 2

3

⇔ C = 60 0 A

B

AB

=

2

AC

+ BC 2 − 2( AC ) ( BC ) cos C

= 10 2 + 16 2 − 2(10) (16) cos 60 0 = 100 + 256 − 160 = 196 = 14 cm C

2. +itu +itun$ n$"a "a! ! ""ua uas s ∆ABC ika 'iketa!ui A% 13,0 & b%16 'an c% 6 cm. Jawab : 16 cm

∆

L ABC

=

1

2

b

b.c

6 cm

⋅ sin sin A

1

⇔ × 16 × 6 × sin sin 135c 0 2

⇔ 48 .

1 2

2

= 24

2

cm

2

a





∆

=

L ABC

∆

=

L ABC

ac

⋅ sin sin B ⋅

AB BC

2

⋅ sin sin B

b

c

1

2

A

= 8 ⋅ sin sin B

4 3

B

2 1

= ⋅ 2 ⋅ 8 ⋅ sin sin B

4 3

sin sin

1

=

B

4 3 8

=

1 2

B

a

3

= 60 0

/. Tentuk entukan an "uas "uas ∆ABC ika 'iketa!ui anan$ a % 6 cm& b% cm 'an c%/ cm * Jawab : S

1

= ( a + b + c) 2 1

⇔ ( 6 + 8 + 4) 2 1

⇔ (18) 2 ⇔ 9 cm

L

=

s ( s

b

c

− a) ⋅ ( s − b ) ⋅ ( s − c )

a

= 9(9 − 6).(9 − 8).(9 − 4) = 9( 3).(1) .(5) = 9 ⋅ 15 = 3 15 cm 2

,. Tentuk entukan an "uas "uas ∆ABC ika 'iketa!ui AB%10 cm 'an BC% cm 'an AC%, cm. C Jawab:







s

= 1 ( AB + BC + AC ) 2

= 1 (10 + 7 + 5) 2 1

= ( 22) = 11 cm 2

∆ ⇔ ⇔ ⇔ ⇔

− AB).( s − BC BC ).( s − AC ) 11(11 − 10).(11 − 7).(11 − 5)

L AB ABC C

=

s ( s

11 (1) ( 4) ( 6) 11 × 24 264

=2

66

cm

2

6. Tentuk entukan an "uas "uas ∆ABC ika 'iketa!ui A%00 'an B%600 'an a% cm. Jawab :

C . +itun$"a! +itun$"a! anan$ anan$ a ika 'iketa!u 'iketa!uana anan$ n$ c%20 cm A%30 A%300 & B%600 * Jawab : b a

∆

L ABC

= =

c

2

sin sin

A

2 sin sin

sin B ⋅ sin

α

20 2 sin sin 30 0 ⋅ sin sin 60 0 2 sin sin (180 − ( 30 + 60) ) 400

C

C

1 1

3

A

b

c

aB





∆

L AB ABC C

50 3

=

= a

2

sin sin

B

⋅ sin sin

2 sin sin

C

A

sin sin 60 ⋅ sin sin (180 − (30 + 60) ) 2 sin sin 30

a

50 3

2

a

1

2

2

=

3 ⋅ sin sin 90 2.

1

1 2

50 3

= a2 ⋅

50 3 1 3 2

= a 2 ⇔ a 2 = 100.3 = 300 ⇔ a =

2

3.1 300

=5

12

cm

. Dik Diketa! eta!ui ui A%1 A%120 200 & B%300 'an C%300 'an anan$ c% 10 cm. +itun$"a! "uas ∆ABC* Jawab :

∆

L ABC

= = =

c

2

sin sin

A

sin B ⋅ sin

2 sin sin

C

10 2 sin sin 120 ⋅ sin sin 30 2 sin sin 30 100 ⋅

1 2

2⋅

= 25

3

3⋅

b

a

1 2

1 c

2

cm

2

. +itun$"a! +itun$"a! anan anan$ $ b& ika ika 'iketa!u 'iketa!uii "uas ∆ABC % 2 cm2& A%/,0 & C% 00 * Jawab : 2





72 =

b

2

sin sin 45 ⋅ sin sin 90

2 sin sin (180 − (45 + 90) ) b

2

⋅

1

b

2

⋅

1

72 =

2 ⋅1 2 2 sin sin 45

72 =

2. 72 2

=

2 ⋅1

2 1

2

2 1

2

2

b

2

⇔ b 2 = 144 ⇔ b =

144

= 12 cm

10 . +itun$"a! "uas ∆ABC& ika 'iketa!ui anan$ sisi a% cm& b% cm 'an c%, cm. Jawab : s

1

= ( a + b + c) 2 1

⇔ ( 7 + 8 + 5) 2 1

b

a

⇔ ( 20) = 10 cm 2

c

= ⇔ ⇔ ⇔

L

− a)( s − b)( s − c) 10(10 − 7)(10 − 8)(10 − 5) s ( s

10(3)(2)(5) 300

=5

12

cm

2

Contoh Soal Luas Segitiga Dalam Trigonometri

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