101 Skills You Need to Survive in the Woods: The Most Effective Wilderness Know-How on Fire-Making

101 Skills You Need to Survive in the Woods: The Most Effective Wilderness Know-How on Fire-Making

Matematika jawab no 1 sampai 5
no ngasal
ngasal report​

jawab no 1 sampai 5
no ngasal
ngasal report​

No. 1

Diketahui :

<B = 90°

<A = 30°

<C = 180° - (90° + 30°) = 60°

panjang BC = 8 cm

Ditanya :

panjang AC = ...?

panjang AB = ...?

Jawab :

* panjang AC

[tex] \frac{ac}{sin \: b} = \frac{bc}{sin \: a} \\ \frac{ac}{sin \: 90} = \frac{8}{sin \: 30} \\ ac \times sin \: 30 = 8 \times sin \: 90 \\ ac \times \frac{1}{2} = 8 \times 1 \\ \frac{ac}{2} = 8 \\ ac = 8 \times 2 \\ ac = 16 \: cm[/tex]

* panjang AB

[tex]ab = \sqrt{ {ac}^{2} - {bc}^{2} } \\ ab = \sqrt{ {16}^{2} - {8}^{2} } \\ ab = \sqrt{256 - 64} \\ ab = \sqrt{192} \\ ab = 8 \sqrt{3} \: cm[/tex]

Jadi, panjang AC dan AB adalah 16 cm dan 83 cm.

No. 2

Diketahui :

<P = 90°

<R = 45°

<Q = 180° - (90° + 45°) = 45°

panjang PQ = 10 cm

Ditanya :

panjang PR = ...?

panjang QR = ...?

Jawab :

* panjang PR

[tex] \frac{pr}{sin \: q} = \frac{pq}{sin \: r} \\ \frac{pr}{sin \: 45} = \frac{10}{sin \: 45} \\ pr \times sin \: 45 = 10 \times sin \: 45 \\ pr \times \frac{1}{ \sqrt{2} } = 10 \times \frac{1}{ \sqrt{2} } \\ \frac{pr}{ \sqrt{2} } = \frac{10}{ \sqrt{2} } \\ pr = \frac{10 \sqrt{2} }{ \sqrt{2} } \\ pr = 10 \: cm[/tex]

* panjang QR

[tex]qr = \sqrt{ {pr}^{2} + {pq}^{2} } \\ qr = \sqrt{ {10}^{2} + {10}^{2} } \\ qr = \sqrt{100 + 100} \\ qr = \sqrt{200} \\ qr = 10 \sqrt{2} \: cm[/tex]

Jadi, panjang PR dan QR adalah 10 cm dan 102 cm.

No. 3

Diketahui :

<E = 90°

<D = 60°

<F = 180° - (90° + 60°) = 30°

panjang EF = 12√3 cm

Ditanya :

panjang DE = ...?

panjang DF = ...?

Jawab :

* panjang DE

[tex] \frac{de}{sin \: f} = \frac{ef}{sin \: d} \\ \frac{de}{sin \: 30} = \frac{12 \sqrt{3} }{sin \: 60} \\ de \times sin \: 60 = 12 \sqrt{3} \times sin \: 30 \\ de \times \frac{ \sqrt{3} }{2} = 12 \sqrt{3} \times \frac{1}{2} \\ \frac{de \sqrt{3} }{2} = \frac{12 \sqrt{3} }{2} \\ de = \frac{12 \sqrt{3} }{ \sqrt{3} } \\ de = 12 \: cm[/tex]

* panjang DF

[tex]df = \sqrt{ {de}^{2} + {ef}^{2} } \\ df = \sqrt{ {12}^{2} + {(12 \sqrt{3}) }^{2} } \\ df = \sqrt{144 + (144 \times 3)} \\ df = \sqrt{144 + 432} \\ df = \sqrt{576} \\ df = 24 \: cm[/tex]

Jadi, panjang DE dan DF adalah 12 cm dan 24 cm.

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