SOLUTION: Acute triangle ABC has \sin A = \dfrac{2}{5} and \sin B = \dfrac{2}{\sqrt{7}}. What is \sin C?

Algebra.Com
Question 1210261: Acute triangle ABC has \sin A = \dfrac{2}{5} and \sin B = \dfrac{2}{\sqrt{7}}. What is \sin C?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52804)   (Show Source): You can put this solution on YOUR website!
.
Acute triangle  ABC  has   sin(A) =   and   sin(B) = .   What is  sin(C)  ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 


Since sin(A) =   and angle A is acute (given), we have

    A =  = 23.5781785 degrees.



Since sin(B) =   and angle B is acute (given), we have

    B =  = arcsin(0.755928946) = 49.1066053 degrees.



Therefore,  C = 180 - 23.5781785 - 49.1066053 = 107.3152162 degrees.


Then  sin(C) = 0.95468179116.


ANSWER.  sin(C)  is about  0.95468179116.

Solved.


----------------------------------

A post-solution note


        As  I  solved the problem to the end,  I see that its formulation
        is  FATALLY  INCORRECT  and is  SELF-CONTRADICTORY.

        Indeed,  the problem says that the given triangle is acute,
        which means that all three its angles are acute.

        But the angle  C,  as  I  found it,  is  OBTUSE:   its measure is about  107  degrees.

So,  this problem simply  DECEIVES  a reader,
which is    for a  Math problem.
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

sin^2(A) + cos^2(A) = 1 ......... pythagorean trig identity
(2/5)^2 + cos^2(A) = 1
4/25 + cos^2(A) = 1
cos^2(A) = 1 - 4/25
cos^2(A) = 21/25
cos(A) = sqrt(21/25)
cos(A) = sqrt(21)/sqrt(25)
cos(A) = sqrt(21)/5

sin^2(B) + cos^2(B) = 1
(2/sqrt(7))^2 + cos^2(B) = 1
4/7 + cos^2(B) = 1
cos^2(B) = 1 - 4/7
cos^2(B) = 3/7
cos(B) = sqrt(3/7)

To summarize so far:
sin(A) = 2/5
cos(A) = sqrt(21)/5
sin(B) = 2/sqrt(7)
cos(B) = sqrt(3/7)

Then,
A+B+C = 180 degrees
C = 180-(A+B)
sin(C) = sin(180 - (A+B) )
sin(C) = sin(A+B) ........................................... use identity sin(180-x) = sin(x)
sin(C) = sin(A)cos(B) + cos(A)sin(B) ......................... angle sum identity
sin(C) = (2/5)*sqrt(3/7) + (sqrt(21)/5)*(2/sqrt(7))
sin(C) = 2*sqrt(3)/(5*sqrt(7)) + 2*sqrt(21)/(5*sqrt(7))
sin(C) = (2*sqrt(3) + 2*sqrt(21))/(5*sqrt(7))
sin(C) = 2*(sqrt(3) + sqrt(21))/(5*sqrt(7)) exactly
sin(C) = 0.9546817913107 approximately

----------------------------------------------------------------------------------------------------------------------

Another approach:
sin(A) = 2/5
A = arcsin(2/5)
A = 23.5781784782 degrees approximately
sin(B) = 2/sqrt(7)
B = arcsin(2/sqrt(7))
B = 49.10660535087 degrees approximately

A+B+C = 180
C = 180-A-B
C = 180-23.5781784782-49.10660535087
C = 107.31521617093
sin(C) = sin(107.31521617093)
sin(C) = 0.9546817913107 approximately
The drawback of this method is that you won't be able to determine the exact form in terms of square roots, as shown in the previous section.
However, some teachers are perfectly fine with the answer in decimal form.

RELATED QUESTIONS

In right triangle ABC,what is the numerical value of sin^2 A + sin^2 B + sin^2 C ? (answered by solver91311)
Triangle ABC has side lengths AB = 8, BC = 10, and AC = 12. Find \dfrac{\cos A}{\cos... (answered by ikleyn,greenestamps)
in a acute angle triangle ABC if SIN (B+C-A) = (SQUARE ROOT 3) /2 and TAN... (answered by ikleyn)
Prove that in an acute angled triangle ABC , a/sin A = b/sin B =c/sin... (answered by robertb)
in ∆ABC, a = 16, b = 21 and sin A= 2/7 find sin... (answered by ewatrrr)
Is an acute angle and sin 0 is given. Use the Pythagorean identity sin^2 + cos^2=1 find (answered by stanbon)
IN Triangle ABC, AB=8, BC=5, and Angle C = 30 degrees. What is sin... (answered by rothauserc)
If angle A and angle B are acute angles of right triangle ABC, what is the value of sin A (answered by jim_thompson5910)
Given: Right Triangle ABC Prove: (sin A)^2 + (sin B)^2 =... (answered by Alan3354)