A problem for discussion-Find sin A?

Question

In a triangle ABC, AB = (4√6)/3, cos B = √(6)/6, and the median BD = √5. Find the exact value of sin A.

This is a challenging problem. I'd like to see different approaches.

Up to now, none of the solutions below is close.

Hint: Draw a segment DE such that E is the mid-point of BC.

breakpara's solution is very close, away by about 2%.

Answer 1

The first step in this problem is to create a point E such that CE forms a right triangle (That is, angle CEB = 90 degrees and angle CEA = 90 degrees).

By the SOHCAHTOA rules,

cosB = adj/hyp, OR
cosB = (BE)/(BC)

sinA = opp/hyp, OR
sinA = (CE)/(AC)

But (BE)^2 + (CE)^2 = (BC)^2 (by the Pythagorean Theorem).
Therefore,
(CE)^2 = (BC)^2 - (BE)^2, and

CE = sqrt ((BC)^2 - (BE)^2)
So

sin(A) = sqrt ((BC)^2 - (BE)^2)/(AC)

Since
cos(B) = (BE)/(BC),
(BC) cos(B) = (BE), so we can replace the value as normal, in the sin(A) equation.

sin(A) = sqrt ((BC)^2 - (BC)^2 [cos(B)]^2 ) / (AC)

Substituting in cos(B) = sqrt(6)/6 and simplifying, we get

sin(A) = sqrt ( (BC)^2 - (BC)^2 [6/36] ) / (AC)
sin(A) = sqrt ( (BC)^2 - (BC)^2 [1/6]) / (AC)

Factoring out (BC)^2 within the square root,

sin(A) = sqrt ( (BC)^2 [1 - 1/6] ) / (AC)
sin(A) = (BC) sqrt (1 - 1/6) / (AC)
sin(A) = {(BC) sqrt (5/6)} / (AC)
sin(A) = sqrt(5/6) * (BC)/(AC)

Answer still in progress....

EDIT: After nearly an hour of trying to get somewhere, I got nowhere. I wish I weren't so sleepy because I really wanted to figure out the problem.

Answer 2

Using a calculator:
B is approximately 65.91 degrees
AB is is approximately 3.27
BD is approximately 2.24

Drawing the problem graphically to scale allows us to determine that A is approximately 36 degrees.

Again using a calculator, sin A is approximately 0.588.

I know this is not the exact solution, but I think it's the best solution so far... I'm going to keep trying to get the exact solution...

Ok... here is your (almost) exact solution:
B = invcos(6^0.5/6) = 65.905157447889 degrees

Draw line AE perpendicular to BC to create a right triangle ABE inside of the original triangle.

BE = AB * cos B = 4/3
AE = AB * sin B = 2.981423970000

Draw line DF perpendicular to BC to create a right triangle BDF inside of the original triangle. Since BD is a median of ABC, DF is exactly 1/2 of AE and EF is exactly 1/2 of EC.

BF = [BD^2 - (AE/2)^2]^(1/2) = 5/3
EF = BF - BE = 1/3
C = invtan [AE/(EF*2)] = 77.395617351622 degrees
A = 180 - B - C = 36.699225200489 degrees

Therefore sin A = 0.597614304667

Ok... let's take this one step further... Let's find the EXACT value of sin A.
AE= (AB^2 - BE^2)^0.5 = (4/3) * 5^0.5
C = invtan[AE/(EF*2)] = invtan (2 * 5^0.5)

Therefore sin A = sin [180 - invcos(6^0.5/6) - invtan (2 * 5^0.5)]
PERFECTION!!!

Answer 3

It is very helpful to draw the triangle in front of you while you read this answer.
In a triangle ABC, with sides a,b,c opposing angles A,B,C, respectively,
The law of sines is a/sinA=b/sinB=c/sinC
The law of cosines is c^2=a^2+b^2-2ab cos(C)

Answer:

In triangle BCD, apply the law of cosines for angle C to get:

5 = a^2+(b/2)^2-2*a*(b/2) cos(C) or
5 = a^2+b^2/4-ab cos(C) or
20 = 4a^2+b^2-4ab cos(C)

Now, applying the law of cosines a second time on the triangle ABC for angle c, we get:

c^2=a^2+b^2-2abcos(C) so

-abcos(C) = (c^2-a^2-b^2)/2 substitute this into the last equation:
20 = 4a^2+b^2+4[(c^2-a^2-b^2)/2] or simplify to reach:

2a^2=b^2-2c^2+20

Now, apply the law of cosine a third time to ABC for angle B:

b^2=a^2+c^2-2ac cos(B) and substitute this for b^2 in the last equation to reach:

2a^2=a^2+c^2-2ac cos(B)-2c^2+20 so

a^2 +2ac cos(B) -(20-c^2) =0

Solve this quadratic equation for a using quadratic formula and note that Cos(B) = sqrt(6)/6, c = 4sqrt(6)/3.

After simplification you get a =[-4+sqrt(172)]/3

Now, b = sqrt(a^2+c^2-2ac cos(B)) to get
b = sqrt[316-32sqrt(43)]/3

Now, sinA/sinB= a/b so
Sin A = sinB (a/b)

sinB = sqrt(1-cos^2(B)) = sqrt(5/6)
a/b = [265+71sqrt(43)]*
sqrt[79-8sqrt(43)]/3489

So SinA = sinB (a/b) = 0.98 approximately.

Answer 4

AB=4(sqrt6)/3
cosB=(sqrt6)/6
BD=sqrt5
hence,sinB=1-cos^2(B)
=(sqrt30)/6
From sine rule,AC/sinB=2R,where R=sqrt5
then,AC=2RsinB=2(sqrt5)(sqrt30)/6
=5(sqrt6)/3
and AB=2RsinC ==> sinC = AB/2R =2(sqrt6)/3(sqrt5)
sinA = sin(180-B-C)=sin(B+C)
=sinBcosC+sinCcosB
=[(sqrt30)/6][sqrt(7/15)]+2(sqrt6)^2/18(sqrt5)
=(sqrt14)/6+2(sqrt15)/15

Answer 5

Synopsis:
Note that the median bisects A: A = A'+A".
Use the law of cosine to figure out AD:
(AD)^2 = (BD)^2 + (AB)^2 - 2 (AB) (BD) Cos(B)
Use law of sine to figure sin(A'):
(BD)/sin(B) = (AD)/sin(A') [with sin(b) =√(1-cos^2(B))]
Use the law of cosine to figure out AC
(AC)^2 =(2 BD)^2 + (AB)^2 - 2 (2 BD) (AB) cos(B)
Use law of sine to figure angle C:
(AB)/sin(C) = (AC)/sin(B)
Use law of sine to figure out angle A":
(BD)/sin(A") = (AD)/sin(C)
Use the trigonometric identity to figure out A
sin(A'+A") = sin(A')cos(A")+cos(A')sin(A")
Now do the work.

Answer 6

Nothing to discuss; just obtuse sine/cosine theorems.