If w = cos (37) - i sin (47) find w^5
without using a calculator.
Answers
Answer:
hkjg
Step-by-step explanation:
the triangle remains the same.
To Find :
Base and Height of the Triangle .
Solution :
\longmapsto\tt{Let\:base\:be(b)=4x}⟼Letbasebe(b)=4x
\longmapsto\tt{Let\:height\:be(h)=3x}⟼Letheightbe(h)=3x
Using Formula :
\longmapsto\tt\boxed{Area\:of\:Triangle=\dfrac{1}{2}\times{b}\times{h}}⟼
AreaofTriangle=
2
1
×b×h
Putting Values :
\longmapsto\tt{\dfrac{1}{{\cancel{2}}}\times{{\cancel{4x}}}\times{3x}}⟼
2
1
×
4x
×3x
\longmapsto\tt{2x\times{3x}}⟼2x×3x
\longmapsto\tt\bf{6x}^{2}⟼6x
2
Now ,
If the base is increased by 4 cm and the height is decreased by 2 cm, the area of the triangle remains the same.
\longmapsto\tt{Base=4x+4}⟼Base=4x+4
\longmapsto\tt{Height=3x-2}⟼Height=3x−2
Using Formula :
\longmapsto\tt\boxed{Area\:of\:Triangle=\dfrac{1}{2}\times{b}\times{h}}⟼
AreaofTriangle=
2
1
×b×h
Putting Values :
\longmapsto\tt{{6x}^{2}=\dfrac{1}{2}\times{(4x+4)}\times{(3x-2)}}⟼6x
2
=
2
1
×(4x+4)×(3x−2)
\longmapsto\tt{6x}^{2}\times{2}=(4x+4)\:\:(3x-2)⟼6x
2
×2=(4x+4)(3x−2)
\longmapsto\tt{{{\cancel{12x}^{2}}}={\cancel{{12x}^{2}}}-8x+12x-8}⟼
12x
2
=
12x
2
−8x+12x−8
\longmapsto\tt{-8x+12x-8=0}⟼−8x+12x−8=0
\longmapsto\tt{4x-8=0}⟼4x−8=0
\longmapsto\tt{4x=8}⟼4x=8
\longmapsto\tt{x=\cancel\dfrac{8}{4}}⟼x=
4
8
\longmapsto\tt\bf{x=2}⟼x=2
Value of x is 2 .
Therefore :
\longmapsto\tt{Base\:of\:Triangle=4(2)}⟼BaseofTriangle=4(2)
\longmapsto\tt\bf{8\:cm}⟼8cm
\longmapsto\tt{Height\:of\:Triangle=3(2)}⟼HeightofTriangle=3(2)
\longmapsto\tt\bf{6\:cm}⟼6cm