Materi : Garis dan Sudut
pemisalan
∠A = a
∠B = 180° - a [ pelurus dari ∠A ]
∠C = 90° - a [ penyiku dari ∠A ]
________________________/
2( ∠B ) + ⅓( ∠C ) = 320°
2( 180° - a ) + ⅓( 90° - a ) = 320°
360° - 2a + 30° - ⅓a = 320°
390° - 320° = 2a + ⅓a
7/3a = 70°
⅓a = 10°
a = 30°
Maka Nilai 5/3 . a = 5/3 . 30° = 50°
Semoga bisa membantu
[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
Jawab:Harusnya jawabannya = 50°
Penjelasan dengan langkah-langkah:Diketahuidua kali pelurus <A = 2(180° - <A)1/3 penyiku <A = (1/3)(90° - <A)
Jumlahnya 320°2(180° - <A) + (1/3)(90° - <A) = 320°360° - 2<A + 30° - (1/3)<A = 320°2<A + (1/3)<A = 360° + 30° - 320°(6/3)<A + (1/3)<A = 70°(7/3)<A = 70°<A = 70° ÷ (7/3)<A = 70° × (3/7)<A = (10 × 3)°<A = 30°
Ditanya (5/3) <A= (5/3) × 30°= 50°
(xcvi)
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Materi : Garis dan Sudut
pemisalan
∠A = a
∠B = 180° - a [ pelurus dari ∠A ]
∠C = 90° - a [ penyiku dari ∠A ]
________________________/
2( ∠B ) + ⅓( ∠C ) = 320°
2( 180° - a ) + ⅓( 90° - a ) = 320°
360° - 2a + 30° - ⅓a = 320°
390° - 320° = 2a + ⅓a
7/3a = 70°
⅓a = 10°
a = 30°
Maka Nilai 5/3 . a = 5/3 . 30° = 50°
Semoga bisa membantu
[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
Jawab:
Harusnya jawabannya = 50°
Penjelasan dengan langkah-langkah:
Diketahui
dua kali pelurus <A = 2(180° - <A)
1/3 penyiku <A = (1/3)(90° - <A)
Jumlahnya 320°
2(180° - <A) + (1/3)(90° - <A) = 320°
360° - 2<A + 30° - (1/3)<A = 320°
2<A + (1/3)<A = 360° + 30° - 320°
(6/3)<A + (1/3)<A = 70°
(7/3)<A = 70°
<A = 70° ÷ (7/3)
<A = 70° × (3/7)
<A = (10 × 3)°
<A = 30°
Ditanya (5/3) <A
= (5/3) × 30°
= 50°
(xcvi)