Asked by Kd
the figure AB = 11cm, BC = 8cm, AD = 3cm, AC = 5cm and DAC is a right angle.
The size of ADC;
The size of ACB;
The area of the quadrilateral ABCD.
tan ADC = 5/3
11^2 = 5^2+8^2 - 2*5*8 cos ACB
Now you can figure the areas of the triangles ACD and ACB, and just add them up for the area of ABCD
DAC = 1/2(AD)(AC)
So area ACB =1/2(AB)(BC)sin(ACB)
The size of ADC;
The size of ACB;
The area of the quadrilateral ABCD.
tan ADC = 5/3
11^2 = 5^2+8^2 - 2*5*8 cos ACB
Now you can figure the areas of the triangles ACD and ACB, and just add them up for the area of ABCD
DAC = 1/2(AD)(AC)
So area ACB =1/2(AB)(BC)sin(ACB)
Answers
Answered by
Reiny
You just repeated the solution Steve gave you here:
https://www.jiskha.com/questions/1765464/In-the-figure-AB-11cm-BC-8cm-AD-3cm-AC-5cm-and-DAC-is-a-right-angle-The
I don't understand why you re-posted the question.
Just do what he told you to do.
https://www.jiskha.com/questions/1765464/In-the-figure-AB-11cm-BC-8cm-AD-3cm-AC-5cm-and-DAC-is-a-right-angle-The
I don't understand why you re-posted the question.
Just do what he told you to do.
Answered by
Anonymous
Try to explain .. homework please
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