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3 years ago

Prove that sin20sin40sin60sin80=3/16

1 answer:

5 0

We'll use the following properties of sine and cosine to prove this:

$sin(a)\cdot sin(b) = \frac12(cos(a-b) - cos(a+b))$

$cos(a)\cdot cos(b) = \frac12(cos(a+b) + cos(a-b))$

$cos(180+a) = cos(180-a)$

Then it's just a matter of filling it in...

sin20sin40 * sin60sin80 = 1/2(cos20 - cos60) * 1/2 (cos20 - cos140) =

1/8( cos40 + 1 - cos160 - cos120 - cos40 - cos80 + cos80 + cos200) =

1/8(1 - cos160 - cos120 + cos200) =

1/8(1 - cos160 - cos120 + cos160) =

1/8(1 - cos120 ) = 1/8( 1 + 1/2 ) = 3/16

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3 years ago

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lys-0071 [83]

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BD is the angle bisector of

AlexFokin [52]

Note: Let us consider, we need to find the $m\angle ABC$ and $m\angle DBC$ .

Given:

In the given figure, BD is the angle bisector of ABC.

To find:

The $m\angle ABC$ and $m\angle DBC$ .

Solution:

BD is the angle bisector of ABC. So,

$m\angle ABD=m\angle DBC$

$3x=x+20$

$3x-x=20$

$2x=20$

Divide both sides by 2.

$x=\dfrac{20}{2}$

$x=10$

Now,

$m\angle DBC=(x+20)^\circ$

$m\angle DBC=(10+20)^\circ$

$m\angle DBC=30^\circ$

And,

$m\angle ABC=(3x)^\circ+(x+20)^\circ$

$m\angle ABC=(4x+20)^\circ$

$m\angle ABC=(4(10)+20)^\circ$

$m\angle ABC=(40+20)^\circ$

$m\angle ABC=60^\circ$

Therefore, $m\angle DBC=30^\circ,m\angle ABD=30^\circ$ and $m\angle ABC=60^\circ$ .

8 0

3 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago

If a right triangle has side lengths of 3 cm and 7 cm, what is the length of the hypotenuse? brainliest will report if answer is

Neko [114]

Answer:

Solution given:

length[a]=3cm

base(b)=7cm

hypotenuse [c]=?

we have

by using Pythagoras law

a²+b²=c²

3²+7²=c²

c=√58=7.61m

the length of the hypotenuse is 7.61 or 7.6 or8m.

8 0

3 years ago