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JeanaShupp

Answer:[tex]cos\ C\times\ sin\ A=\frac{\sqrt{3} }{2}\times\ \frac{\sqrt{3} }{2}=\frac{3}{4}[/tex]


Step-by-step explanation:

Given a right triangle ABC where AB = 1, AC = 2

As it is a right triangle, thus by Pythagoras theorem

[tex]AC^2=AB^2+BC^2\\\Rightarrow2^2=1^2+BC^2\\\Rightarrow4=1+BC^2\\\Rightarrow\ BC^2=4-1=3\\\Rightarrow\ BC=\sqrt{3}[/tex]

Now ,

as [tex]sin A=\frac{sides\ opposite\ to\ angle\ A }{hypotenuse}[/tex]

[tex]sin\ A=\frac{BC}{AC}=\frac{\sqrt{3} }{2}[/tex]

and

[tex]cos C=\frac{side\ adjacent\ to\ angle\ C}{hypotenuse}[/tex]

[tex]cos\ C=\frac{BC}{AC}=\frac{\sqrt{3} }{2}[/tex]


Therefore,

[tex]cos\ C\times\ sin\ A=\frac{\sqrt{3} }{2}\times\ \frac{\sqrt{3} }{2}=\frac{3}{4}[/tex]

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Answer:Cos C × Sin A =  [tex]\frac{3}{4}[/tex] and BC = √3.

Step-by-step explanation:

Given :  A right triangle ABC where AB = 1, AC = 2 .

To Find  : BC and cos C × sin A .

Solution :We have ΔABC is a right triangle, AB = 1, AC = 2.

As it is a right triangle, thus by Pythagoras theorem

AC² = AB² +BC².

2² = 1² + BC²

4 = 1 + BC²

BC = √3.

Now ,

Sin A = [tex]\frac{side\ opposite\ of\ angle\ A}{Hypontnuse}[/tex]

Sin A = [tex]\frac{BC}{AC}[/tex].

Sin A = [tex]\frac{√3}{2}[/tex].

Cos C =  [tex]\frac{side\ opposite\ of\ angle\ C}{Hypontnuse}[/tex].

Cos C = [tex]\frac{BC}{AC}[/tex].

Cos C = = [tex]\frac{√3}{2}[/tex].

Cos C × Sin A

[tex]\frac{√3}{2}[/tex] ×[tex]\frac{√3}{2}[/tex].

⇒ [tex]\frac{3}{4}[/tex].

Therefore, Cos C × Sin A =  [tex]\frac{3}{4}[/tex] and BC = √3.

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