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2 years ago

An equilateral triangle has a semiperimeter of 6 meters.

Mathematics

2 answers:

Ierofanga [76]2 years ago

8 0

Answer:

Step-by-step explanation:

Took the test

guapka [62]2 years ago

7 0

Answer:

Step-by-step explanation:

An equilateral triangle has a semiperimeter of 6 meters.

Heron’s formula: Area = StartRoot s (s minus a) (s minus b) (s minus c) EndRoot

An equilateral triangle has a semiperimeter of 6 meters. What is the area of the triangle? Round to the nearest square meter.

2 square meters

7 square meters

20 square meters

78 square meters

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$\tan A=\dfrac{\sin A}{\sin C}$

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$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

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By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$

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Option B is false.

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By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

Now

$\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.$

Option C is false.

$\cos A=\tan C.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

As you can see $\cos A\neq \tan C$ and option D is not true.

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By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

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