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Yanka [14]
3 years ago
12

Fine length of BC on the following photo.

Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

BC=4\sqrt{5}\ units

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

AC^2=AD^2+DC^2

substitute the given values

AC^2=16^2+8^2

AC^2=320

AC=\sqrt{320}\ units

simplify

AC=8\sqrt{5}\ units

step 2

In the right triangle ACD

Find the cosine of angle CAD

cos(\angle CAD)=\frac{AD}{AC}

substitute the given values

cos(\angle CAD)=\frac{16}{8\sqrt{5}}

cos(\angle CAD)=\frac{2}{\sqrt{5}} ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

cos(\angle BAC)=\frac{AC}{AB}

substitute the given values

cos(\angle BAC)=\frac{8\sqrt{5}}{16+x} ----> equation B

step 4

Find the value of x

In this problem

\angle CAD=\angle BAC ----> is the same angle

so

equate equation A and equation B

\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}

solve for x

Multiply in cross

(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units

DB=4\ units

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

BC^2=DC^2+DB^2

substitute the given values

BC^2=8^2+4^2

BC^2=80

BC=\sqrt{80}\ units

simplify

BC=4\sqrt{5}\ units

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