In a circuit electrons flow in one direction.
2 answers:
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Step-by-step explanation:
The ratios are;
\dfrac{BC}{AB} = \dfrac{3}{5}
AB
BC
=
5
3
\dfrac{AC}{AB} = \dfrac{4}{5}
AB
AC
=
5
4
\dfrac{BC}{AC} = \dfrac{3}{4}
AC
BC
=
4
3
\dfrac{DE}{AD} = \dfrac{3}{5}
AD
DE
=
5
3
\dfrac{AE}{AD} = \dfrac{4}{5}
AD
AE
=
5
4
\dfrac{DE}{AE} =\dfrac{3}{4}
AE
DE
=
4
3
koGiven that the lengths of the sides are;
\overline {AB}
AB
= 20
\overline {BC}
BC
= 12
\overline {AC}
AC
= 16
\overline {AD}
AD
= 10
\overline {DE}
DE
= 6
\overline {AE}
AE
= 8
The ratios are;
\dfrac{Length \ opposite \ \angle A}{Hypothenus} = \dfrac{BC}{AB} = \dfrac{12}{20} = \dfrac{3}{5}
Hypothenus
Length opposite ∠A
=
AB
BC
=
20
12
=
5
3
\dfrac{Length \ adjacent\ \angle A}{Hypothenus} = \dfrac{AC}{AB} = \dfrac{16}{20} = \dfrac{4}{5}
Hypothenus
Length adjacent ∠A
=
AB
AC
=
20
16
=
5
4
\dfrac{Length \ opposite \ \angle A}{Length \ adjacent \ \angle A} = \dfrac{BC}{AC} = \dfrac{12}{16} = \dfrac{3}{4}
Length adjacent ∠A
Length opposite ∠A
=
AC
BC
=
16
12
=
4
3
\dfrac{Length \ opposite \ \angle A}{Hypothenus} = \dfrac{DE}{AD} = \dfrac{6}{10} = \dfrac{3}{5}
Hypothenus
Length opposite ∠A
=
AD
DE
=
10
6
=
5
3
\dfrac{Length \ adjacent\ \angle A}{Hypothenus} = \dfrac{AE}{AD} = \dfrac{8}{10} = \dfrac{4}{5}
Hypothenus
Length adjacent ∠A
=
AD
AE
=
10
8
=
5
4
\dfrac{Length \ opposite \ \angle A}{Length \ adjacent \ \angle A} = \dfrac{DE}{AE} = \dfrac{6}{8} = \dfrac{3}{4}
Length adjacent ∠A
Length opposite ∠A
=
AE
DE
=
8
6
=
4
3
