How do you find #angleB# if in triangle ABC, #a = 15#, #b = 20# , and #angle A=30^@#?

Answer 1

The answer is: #/_B = sin^-1((20*sin(30^∘))/15)#

You have to use the law of sines, just as you thought. This law is: #a/(sin/_A) = b/(sin/_B) = c/(sin/_C)# We only need the first equation here, let's fill in the numbers. #15/sin(30^∘) =20/(sin/_B)# We use cross-multiplication, then we find: #sin/_B = (20*sin(30^∘))/15# then by taking the inverse sine, we find #/_B = sin^-1((20*sin(30^∘))/15)#

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Answer 2

Oliver Atkinson

To find angle B in triangle ABC, where side a is 15 units long, side b is 20 units long, and angle A is 30 degrees, you can use the Law of Sines or the Law of Cosines.

Using the Law of Cosines:

Use the Law of Cosines to find angle C: ( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) ).
Substitute the known values: ( c^2 = 15^2 + 20^2 - 2(15)(20) \cdot \cos(C) ).
Solve for ( c ).
Use the Law of Sines to find angle B: ( \frac{\sin(B)}{b} = \frac{\sin(A)}{a} ).
Substitute the known values and the calculated value of side c.
Solve for angle B.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.