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In triangle ABC, AB = BC = 25 and AC = 40. What is sin angle ACB?

 Sep 29, 2020
 #1
avatar+27847 
+1

Draw yourself a picture of this isosceles triangle.....you need to find the height by dropping a line from the apex to the base to form a right triangle.....then you can calculate height:

h^2 + (1/2b)^2 = 25^2

h ^2 = 25^2 - (1/2 (40))^2

h = 15

 

Now you can calculate sine ACB = opposite/hypotenuse =   h/25 = 15/25 = 3/5 = .6

 Sep 29, 2020
 #4
avatar+216 
-2

Thank you so much!

Noori  Sep 29, 2020
 #2
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Angle ACB = arccos(20 / 25) = 36.86989765 degrees

 Sep 29, 2020
 #3
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AC/2 = 20      BC = 25          ∠ACB = arccos ((AC/2) / BC)

Guest Sep 29, 2020

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