Nomor 1 dan caranya ya:).... Question from @vivian27

Risep B.
$\sqrt{20^{2}+ 15^{2} } \\ =\sqrt{400+225} \\ = \sqrt{625} \\ =25~cm$

a. Sebangun berarti sudut sudut pada segitiga ABC₁ sama dengan sudut sudut pada segitiga C₂DE,
∠A = ∠C₂
∠B = ∠D
∠C₁ = ∠E

$sin~C1= \frac{AB}{AC}= \frac{20}{25}= \frac{4}{5} \\ C1=53^{0}$
∠E=180-90-37=53°

∠C₂=180 - 53 - 90 = 37°
∠A=180 - 53 - 90= 37°

∠B = ∠D = 90°

(Terbukti sebangun)

c)
$sin~E= \frac{DC}{CE} \\ \\ Sin~53^{o}= \frac{15}{CE} \\ \\ 0,8= \frac{15}{CE} \\ \\ CE=18,75~cm \\ \\ \\ Sin~C_{2} = \frac{DE}{EC} \\ \\ Sin~37^{o}= \frac{DE}{18,75} \\ \\ 0,6=\frac{DE}{18,75} \\ \\ DE=11,25~cm$

1 votes Thanks 1

Naufalmuafa B. kita gunakan phitagoras :
   $\sqrt{ AB^{2} + BC^{2} }$

   $\sqrt{ 20^{2} + 15^{2} }$

   $\sqrt{400+225}$

   $\sqrt{625}$

25cm

c. CE :

$\frac{AB}{CD} = \frac{AC}{CE}$

$\frac{20}{15} = \frac{25}{CE}$

$\frac{4}{3} = \frac{25}{CE}$

CE= (25÷4)×3
CE= 6,25×3
CE= 18,75cm

DE :

$\frac{AB}{CD} = \frac{BC}{DE}$

$\frac{20}{15} = \frac{15}{DE}$

$\frac{4}{3} = \frac{15}{DE}$

DE= (15÷4)×3
DE= 3,75×3
DE=11,25cm

yang A saya bingung

0 votes Thanks 1

Tolong no 9 dan cara :) nya yaaaa

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