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avatar+57 

Find AC in the diagram to the nearest integer.

 

 

\(In $\triangle PQR,$ we have $\angle P = 90^\circ,$ $PQ=3,$ and $QR=\sqrt {58}.$ Find $\tan R.$ \)

 

\(Find the value of $\sin 40^\circ \cdot \cos 50^\circ + \cos 40^\circ \cdot \sin 50^\circ.$\)(Says: "Find the value of...")

 

In the diagram below, we know \(\tan \theta = \frac{3}{4}. \) Find the area of the triangle.

 



 

 May 30, 2019
 #1
avatar+57 
+1

Nvm, the last one I got it. It is 720

 May 30, 2019
 #2
avatar+8873 
+3

Find AC in the diagram to the nearest integer.

 

sin( angle )  =  opposite / hypotenuse

 

sin( 70° )  =  31 / AC

                                         Multiply both sides of the equation by  AC

AC sin( 70° )  =  31

                                         Divide both sides of the equation by  sin( 70° )

AC  =  31 / sin( 70° )

                                         Plug    31 / sin( 70° )    into a calculator.

AC  ≈  33

 May 30, 2019
 #3
avatar+8873 
+3

In \(\triangle PQR,\) we have \(\angle P = 90^\circ,\) \(PQ=3,\) and \(QR=\sqrt {58}.\) Find \(\tan R.\)

 

 

By the Pythagorean theorem,

32 + ( PR )2  =  √[ 58 ]2

9  +  ( PR )2  =  58

( PR )2  =  49

PR  =  7

 

tan( angle )  =  opposite / adjacent

tan( R )  =  3 / PR

tan( R )  =  3 / 7

 May 30, 2019
 #4
avatar+8873 
+3

Find the value of   \(\sin40°\cdot\cos50°+\cos40°\cdot\sin50°\) .

 

Remember the angle addition formula for sine:    \(\sin(\alpha+\beta)\ =\ \sin\alpha\cos\beta+\cos\alpha\sin\beta\)

So...

 

\(\sin40°\cdot\cos50°+\cos40°\cdot\sin50°\ =\ \sin(40°+50°)\ =\ \sin(90°)\ =\ 1\)

_ _ _ _ _

 May 30, 2019
 #5
avatar+8873 
+3

Part of this last problem is just like the second problem.......

 

In the diagram below, we know  \(\tan\theta=\frac34\) . Find the area of the triangle.

 

Let the base of the triangle be the side with length 60. Draw a height to that base and call it  h .

 

 

By the Pythagorean Identity,

1+ cot2θ  =  csc2θ

 

 

1 + ( 4/3 )2  =  1 / sin2θ

 

25 / 9  =  1 / sin2θ

 

 

9 / 25  =  sin2θ

 

sin θ  =  3 / 5

 

 

 

sin( angle )  =  opposite / hypotenuse 
sin θ  =  h / 40

 

 

3 / 5  =  h / 40

 

24  =  h

 

 

 

area of triangle  =  (1/2)(base)(height) 

area of triangle  =  (1/2)(60)(24)

 

 

area of triangle  =  720

 

 May 30, 2019

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