Minta tolong dong kak scptnya

Minta tolong dong kak scptnya

Di atas adalah penampakan segitiga no. 1

Pertama-tama mencari AC dengan rumus phytagoras, karena segitiga tersebut siku siku, maka berlaku rumus:

${a}^{2} + {b}^{2} = {c}^{2} \ {pq}^{2} + {bc}^{2} = {ac}^{2} \ {9}^{2} + {12}^{2} = {ac}^{2} \ {ac}^{2} = 81 + 144 \ ac = sqrt{225} \ ac = 15$

Jadi

1.a.

$sin : a = frac{depan}{miring} = frac{cb}{ac} = frac{12}{15} = frac{4}{5}$

1.b

$cos : a = frac{samping}{miring} = frac{ab}{ac} = frac{9}{15} = frac{3}{5}$

1.c

$tan : a = frac{depan}{samping} = frac{cb}{ab} = frac{12}{9} = frac{4}{3}$

1.d

$sin : c = frac{depan}{miring} = frac{ab}{ac} = frac{9}{15} = frac{3}{5}$

1.e

$cos : c = frac{samping}{miring} = frac{cb}{ac} = frac{12}{15} = frac{4}{5}$

1.f

$tan : c = frac{depan}{samping} = frac{ab}{cb} = frac{9}{12} = frac{3}{4}$

2. Pada no. 2, karena segitiga siku-siku, maka berlaku juga rumus phytagoras dan pada kuadran 3, hanya ( Tan ) yang positif "+"

${c}^{2} = {a}^{2} + {b}^{2} \ {pr}^{2} = {qr}^{2} + {qp}^{2} \ {qr}^{2} = {pr}^{2} - {qp}^{2} \ {qr}^{2} = {25}^{2} - {7}^{2} \ {qr}^{2} = 625 - 49 \ qr = sqrt{576} \ qr = 24$

2.a

– $sin : p = - frac{depan}{miring} = - frac{qr}{pr} = - frac{24}{25}$

2.b

– $cos : p = - frac{samping}{miring} = - frac{pq}{pr} = - frac{7}{25}$

2.c

$tan : p = frac{depan}{samping} = frac{qr}{pq} = frac{24}{7}$

2.d

– $sin : r = -frac{depan}{miring} = - frac{pq}{pr} = - frac{7}{25}$

2.e

– $cos : r = - frac{samping}{miring} = - frac{qr}{pr} = - frac{24}{25}$

2.f

$tan : r = frac{depan}{samping} = frac{pq}{pr} = frac{7}{24}$

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