Question

ΔABC is a right triangle in which  is a right angle, AB = 1, AC = 2, and BC = .

cos C × sin A =

Answer 1

Answer:Cos C × Sin A =
`(3)/(4)` and BC = √3.

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 =
`(side\ opposite\ of\ angle\ A)/(Hypontnuse)`

Sin A =
`(BC)/(AC)`.

Sin A =
`(√3)/(2)`.

Cos C =
`(side\ opposite\ of\ angle\ C)/(Hypontnuse)`.

Cos C =
`(BC)/(AC)`.

Cos C = =
`(√3)/(2)`.

Cos C × Sin A

`(√3)/(2)` ×
`(√3)/(2)`.

⇒
`(3)/(4)`.

Therefore, Cos C × Sin A =
`(3)/(4)` and BC = √3.

Answer 2

Answer:
`cos\ C*\ sin\ A=(√(3) )/(2)*\ (√(3) )/(2)=(3)/(4)`

Explanation:

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

As it is a right triangle, thus by Pythagoras theorem

`AC^2=AB^2+BC^2\\\Rightarrow2^2=1^2+BC^2\\\Rightarrow4=1+BC^2\\\Rightarrow\ BC^2=4-1=3\\\Rightarrow\ BC=√(3)`

Now ,

as
`sin A=(sides\ opposite\ to\ angle\ A )/(hypotenuse)`

`sin\ A=(BC)/(AC)=(√(3) )/(2)`

and

`cos C=(side\ adjacent\ to\ angle\ C)/(hypotenuse)`

`cos\ C=(BC)/(AC)=(√(3) )/(2)`

Therefore,

`cos\ C*\ sin\ A=(√(3) )/(2)*\ (√(3) )/(2)=(3)/(4)`