Pls help me with this question it's due today
2 answers:
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See the picture attached to better understand the problem
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer is
the hypotenuse is (3√2)/2
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer is
the hypotenuse is (3√2)/2
Please see the attached image for a visual representation of our scale factor. We can set up this proportion by taking the DVD cover and poster values and placing them in fractions. Cross multiply and divide to solve for x.
