Question 167754This question is from textbook Algebra 1
: Hello! My problem is a triangle with a measure of an acute angle A which is 29 degrees, side CA of 18 meters, and a right angle which is angle C. Here's what I found out so far:
that angle b = 61 degrees
but I do not know how to figure out CB and BA
if you could help, please show step by step instuctions to help me.
Thank You!
This question is from textbook Algebra 1
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A which is 29 degrees,
side CA of 18 meters,
and a right angle which is angle C.
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Here's what I found out so far:
that angle b = 61 degrees
but I do not know how to figure out CB and BA
--------------------
Draw the picture.
CB is opposite angle A=29 degrees.
BA is opposite angle C so BA is the hypotenuse
CA is opposite angle B
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So, cos(A) = adjacent/hypotenuse
cos(29) = 18/hypotenuse
hypotenuse = 18/cos(29) = 20.58
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sin(A) = CB/hypotenuse
CB = hypotenuse*sin(A)
CB = 20.58*0.4848..
CB = 9.97738....
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Cheers,
Stan H.
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