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# Triangle question

0
234
3

In triangle ABC

Right at B

AB=35

AC=50

Find CB

and find Angle C

Jun 11, 2019

### Best Answer

#1
+8966
+3

By the Pythagorean Theorem,

352 + ( CB )2  =  502

Subtract  352  from both sides of the equation.

( CB )2  =  502 - 352

Simplify the right side of the equation.

( CB )2  =  1275

Since  CB  is a length, take the positive square root of both sides.

CB  =  √[ 1275 ]

And we can simplify the radical.

CB  =  5√[ 51 ]

To get an approximate solution, plug it into a calculator.

CB  ≈  35.707

sin( angle )  =  opposite / hypotenuse

sin( C )  =  35 / 50

Take the inverse sine of both sides of the equation.

C  =  arcsin( 35 / 50 )

Plug  arcsin(35/50)  into a calculator to get...

C  ≈  44.427°

Jun 11, 2019

### 3+0 Answers

#1
+8966
+3
Best Answer

By the Pythagorean Theorem,

352 + ( CB )2  =  502

Subtract  352  from both sides of the equation.

( CB )2  =  502 - 352

Simplify the right side of the equation.

( CB )2  =  1275

Since  CB  is a length, take the positive square root of both sides.

CB  =  √[ 1275 ]

And we can simplify the radical.

CB  =  5√[ 51 ]

To get an approximate solution, plug it into a calculator.

CB  ≈  35.707

sin( angle )  =  opposite / hypotenuse

sin( C )  =  35 / 50

Take the inverse sine of both sides of the equation.

C  =  arcsin( 35 / 50 )

Plug  arcsin(35/50)  into a calculator to get...

C  ≈  44.427°

hectictar Jun 11, 2019
#2
+2

Thank you!

Guest Jun 11, 2019
#3
+2

Oh that was a simple and small great solution!!

I actually did a long one using the Law of cosines

Since I have all three side lengths

therefore AB^2=AC^2 + CB^2 -2AC*BC*Cos(angle c)

i also got to the result 44.4

Guest Jun 11, 2019