3 years ago

If A+B+C=180 Prove that Sin2A+Sin2B+Sin2C=4sinASinBSinC

1 answer:

3 0

73.246.170.19673.246.170.196

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yKpoI14uk [10]

Answer:

No. They are not in proportion

Step-by-step explanation:

$\sf \dfrac{4}{5} \ \text{is in the simplest form}$

$\sf \text{Simplest form of $\dfrac{12}{20}$} = \dfrac{12 \div 4}{20 \div 4}=\dfrac{3}{5}$

So, they are not in proportion

7 0

2 years ago

Marrrta [24]

Answer:

x ^2 − 2 y ^2 −8 x y

Step-by-step explanation:

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2 years ago

MAVERICK [17]

The answer will turn out to be 26.87

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3 years ago

sweet-ann [11.9K]

56% = 14 questions

1% = 14 ÷ 56 = 0.25 questions

100% = 0.25 x 100 = 25 questions

Answer: 25 questions.

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3 years ago

Alja [10]

Answer:

the correct answer is 21.07070206.

6 0

3 years ago

If A+B+C=180 Prove that Sin2A+Sin2B+Sin2C=4sinASinBSinC​