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In a parallelogram ABCD, AB=DC, and AD=CB. The diagonal BD=20cm and the diagonal AC=14cm. If angle ACB=85 degrees, calculate the sizes of the two interior angles of the parallelogram.

 Dec 29, 2014

Best Answer 

 #3
avatar+23245 
+5

Call the intersection point of the two diagonals X.

Looking at Triangle BCX:  ∠BCX =  85°,  BX  =  10  and  CX  =  7.

Law of Sines for Triangle BCX:  sin(∠BCX) / BX  =  sin(∠CBX) / CX

                        --->               sin(85°) / 10  =  sin(∠CBX) / 7

                        --->               7 x sin(85°) / 10   =  sin(∠CBX)

1)  Find ∠CBX.

2)  Use ∠CBX and ∠BCX to find ∠BXC.

3)  Use ∠BXC to find ∠BXA.

4)  Use the Law of Cosines on Triangle AXB to find AB.

5)  Use the Law of Sines on Triangle AXB to find ∠ABX.

6)  Find ∠ABC.

7)  Find ∠BAD.

 Dec 29, 2014
 #1
avatar+360 
0

Oh gawd

 Dec 29, 2014
 #2
avatar
0

I posted this and I can't do it either

 Dec 29, 2014
 #3
avatar+23245 
+5
Best Answer

Call the intersection point of the two diagonals X.

Looking at Triangle BCX:  ∠BCX =  85°,  BX  =  10  and  CX  =  7.

Law of Sines for Triangle BCX:  sin(∠BCX) / BX  =  sin(∠CBX) / CX

                        --->               sin(85°) / 10  =  sin(∠CBX) / 7

                        --->               7 x sin(85°) / 10   =  sin(∠CBX)

1)  Find ∠CBX.

2)  Use ∠CBX and ∠BCX to find ∠BXC.

3)  Use ∠BXC to find ∠BXA.

4)  Use the Law of Cosines on Triangle AXB to find AB.

5)  Use the Law of Sines on Triangle AXB to find ∠ABX.

6)  Find ∠ABC.

7)  Find ∠BAD.

geno3141 Dec 29, 2014
 #4
avatar+128089 
0

Nice one, geno...!!!!

 

 Dec 30, 2014

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