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2 years ago

I’m lost please help

Mathematics

1 answer:

vagabundo [1.1K]2 years ago

5 0

Answer:

See proof below

Step-by-step explanation:

Two triangles are said to be congruent if one of the 4 following rules is valid

The three sides are equal
The three angles are equal
Two angles are the same and a corresponding side is the same
Two sides are equal and the angle between the two sides is equal

Let's consider the two triangles ΔABC and ΔAED.

ΔABC sides are AB, BC and AC

ΔAED sides are AD, AE and ED

We have AE = AC and EB = CD

So AE + EB = AC + CD

But AE + EB = AB and AC+CD = AD

We have

AB of ΔABC = AD of ΔAED

AC of ΔABC = AE of ΔAED

Thus two sides the these two triangles. In order to prove that the triangles are congruent by rule 4, we have to prove that the angle between the sides is also equal. We see that the common angle is ∡BAC = ∡EAC

So triangles ΔABC and ΔAED are congruent

That means all 3 sides of these triangles are equal as well as all the angles

Since BC is the third side of ΔABC and ED the third side of ΔAED, it follows that

BC = ED Proved

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Find f. A) 7.4 B) 8.2 C) 10.5 D) 11.1

tatyana61 [14]

F=72

g=6

------------

$\cos { \left( F \right) } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }$

Therefore:

$\cos { \left( 72 \right) } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right) } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }$

$\\ \\ 12e\cdot \cos { \left( 72 \right) } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right) } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 } } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right) } \right) +36 }$

But what is e?

E=76

G=32

g=6

And:

$\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } }$

Which means that:

$\frac { e }{ \sin { \left( 76 \right) } } =\frac { 6 }{ \sin { \left( 32 \right) } } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } }$

If you take this value into account, you will discover that f is...

$f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } \left( \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } -12\cos { \left( 72 \right) } \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)$

So I would have to say that the answer is approximately (c).

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Answer:

Step-by-step explanation:

I hope this helps!!

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Step-by-step explanation:

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