• Nov 12th 2009, 07:46 PM
Vicky1997
Quadrilateral ABCD has AB = BC = CD. Angle ABC = 70 degrees and Angle BCD = 170 degrees. What is the degree measure of angle BAD?

Could I just get some hints please?

Vicky.
• Nov 12th 2009, 10:07 PM
Wilmer
Let AB = BC = CD = 1

Use Sine Law and isosceles triangle BCD to calculate BD (you should get ~1.992)

Do you see that angle ABD = 65 (70 - 5) ?
Use Cosine Law and triangle ABD to calculate AD (you should get ~1.813)

We now have BD / sin(A) = AD / sin(65)
From that, calculate angle A (you should get 85 degrees.

If you draw a diagram, it'll be easier for you to follow above steps.
• Nov 14th 2009, 10:22 AM
Vicky1997
Quote:

Originally Posted by Wilmer
Let AB = BC = CD = 1

Use Sine Law and isosceles triangle BCD to calculate BD (you should get ~1.992)

Do you see that angle ABD = 65 (70 - 5) ?
Use Cosine Law and triangle ABD to calculate AD (you should get ~1.813)

We now have BD / sin(A) = AD / sin(65)
From that, calculate angle A (you should get 85 degrees.

If you draw a diagram, it'll be easier for you to follow above steps.

Thanks for your help.