Question

1stangle(∠DBE)=(x+3)
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\sf{2nd\: angle \: (\angle{DBC}) = {(x + 20)}^\circ}2ndangle(∠DBC)=(x+20)
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\sf{3rd\: angle\: (\angle{CBF}) = {(x + 7)}^\circ}3rdangle(∠CBF)=(x+7)
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To find:-
\sf{Value\: of\: x }Valueofx
Solution:-
\sf{\angle{DBE} + \angle{DBC} + \angle{CBF} = {180}^\circ\:\:\:\:(Linear\: Pair)}∠DBE+∠DBC+∠CBF=180
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(LinearPair)

= \sf{{(x + 3)}^\circ + {(x + 20)}^\circ + {(x + 7)}^\circ = {180}^\circ}(x+3)
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+(x+20)
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+(x+7)
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=180
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=> \sf{x + {3}^\circ + x + {20}^\circ + x + {7}^\circ = {180}^\circ}x+3
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+x+20
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+x+7
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=180
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=> \sf{3x + {30}^\circ = {180}^\circ}3x+30
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=180
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=> \sf{3x = {180}^\circ - {30}^\circ}3x=180
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−30
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=> \sf{3x = {150}^\circ}3x=150
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=> \sf{x = \dfrac{150^\circ}{3}}x=
3
150
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​


=> \sf{x = {50}^\circ}x=50
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\sf{\therefore\: The\: value\: of \: x \: is\: {50}^\circ}∴Thevalueofxis50
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Now,

The measure of each angle:-

\sf{\angle{DBE} = {(x + 3)}^\circ = {(50 + 3)}^\circ = {53}^\circ}∠DBE=(x+3)
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=(50+3)
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=53
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\sf{\angle{DBC} = {(x + 20)}^\circ ={( 50 + 20)}^\circ = {70}^\circ}∠DBC=(x+20)
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=(50+20)
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=70
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\sf{\angle{CBF} = {(x + 7)}^\circ = {(50 + 7)}^\circ = {57}^\circ}∠CBF=(x+7)
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=(50+7)
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=57
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Answer 1

Answer:

you answer is 57°

siso hope u loved it

Answer 2

Explanation:

What is cos3x equal to? How can you prove it?

.

Cos(3x)=4Cos3(x)−3Cos(x)

Cos(3x)=Cos(2x+x)=Cos(2x)Cos(x)−Sin(2x)Sin(x)

=[2Cos2(x)−1]Cos(x)−2Sin(x)Cos(x)Sin(x)

=2Cos3(x)−Cos(x)−2Cos(x)[Sin2(x)]

=2Cos3(x)−Cos(x)−2Cos(x)[1−Cos2(x)]

=2Cos3(x)−Cos(x)−2Cos(x)+2Cos3(x)

=4Cos3(x)−3Cos(x)

See, I don’t know how to use the [maths] stuff..

I tried to write it the best way.

Basically, put 3x=2x+x and use the identity Cos(A+B)

After that, using other identities of Cos(2A) and Sin(2A) , convert all terms into cos. Open the brackets, add, subtract, and voilà, you got it!!