Advanced algebra and Trig?

Question

How do you find angle B in a triangle is the two sides a = 24 and b = 47 are give along with angle A = 36 degrees.

Answer 1

Use the Law of Sines.

sin A/a = sin B/b
(sin 36)/24 = sin B/47
sin B = (47 sin 36)/24
sin B = 1.151079452
At this point you have a problem--it is impossible for the sine of an angle to have a value greater than 1.

So either you copied the problem wrong, or you are being asked to recognize that a triangle cannot be formed with these dimensions.

Answer 2

First draw a triangle ABC where AB is at the bottom and C is at the top.
Assign AB = c, BC = a, CA = b.

First, draw altitude CD(D is now on AB). You can calculate the following:
sin = opp./hyp.
sin A = CD/CA or sin A = CD/b
sin B = CD/CB or sin B= CD/a

Solving for CD in both equations:
CD = b sin A
CD = a sin B

Using transitive property:
b sin A = a sin B

Divide by sin A sin B:
a/sin A = b/sin B

In fact, you can use other altitudes and you'll acquire the Sine Law:
a/sin A = b/sin B = c/sin C

But for your problem we will only use the first 2 members of the Law.
a/sin A = b/sin B

Now,
a = 24
b = 47
A = 36
B = ?

But before you calculate, you must first check your data.

You are given 2 sides and an angle opposite one of the sides. This case is sometimes ambiguous, because in Geometry, we only have SSS, SAS, ASA postulates and AAS theorem- we do not have SSA theorem/postulate. Follow these rules:

1)
If A > 90 or A = 90, and
a) a > b, then 1 triangle is created,
b) a = b or a < b, then no triangle is created

2)
If A < 90, and
a) a > b, then 1 triangle is created
b) a = b, then 1 triangle is created
c) a < b, then
--- c) 1) 2 triangles is created if a > b sin A
--- c) 2) 1 triangle is created if a = b sin A
--- c) 3) no triangle is created if a < b sin A

Your example applies rule c) 3) so no triangle is formed and angle B does not exist. If you compute it, it will yield an irreal number
B = arcsin(47 sin 36/24)

^_^

Answer 3

Without a picture showing where the angles A and B are you have to explain where they are located relative to the two sides a and b. Is angle A in between the two sides a and b? And which of the remaining two angles is the one labeled B?

I use the law of cosines to solve most triangle problems. It's like the Pythagorean Theorem with a correction term stuck in. I remember it as "one side squared equals the sum of the squares of the other two sides minus two times the product of those same two sides times the cosine of the angle between them."

[By the way, I find it hard to believe that this problem is unsolvable as a couple of the other responders claim.]

Answer 4

Law of Sines:
(sin A) / a = (sin B) / b
(sin 36) / 24 = (sin B) / 47
Multiplying both sides by 47,
(sin B) = 47(sin 36) / 24
Then use the inverse sine to find B.
Unfortuantely, this value is greater than one, so there is no inverse sine.

Your problem is unsolvable.

Answer 5

well you can use law of sin's
which is sin A / a = sin B / b = sin C / c

so we have Sin (36) / 24 = Sin (B) / 47

B = arcsin ( (47 x sin 36) /24))
actually this is not possible because (47 x sin 36) /24) = 1.150.....

and we cannot take an arcsin of a number greater than 1

Answer 6

Sine Law

a/sin A = b/sin B

or sin B = b*sin A/a = 47*sin(36)/24 = 0.796525

Now B = asin(0.796525)

Answer 7

1 . try finding a square plus b square = csquare .

2. then find the square root of 24 square then the square root of 47 square.

3. then and or subtract to find c square!! you should be fine!!

Answer 8

Lets name the unknown side x,
x² = a² + b² -2*a*b*cosA

Answer 9

first find side c then use the cos rule in trig

Answer 10

you are going to have to use the law of cosines to solve for this triangle