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### Course: Trigonometry>Unit 4

Lesson 6: Challenging trigonometry problems

# Trig challenge problem: arithmetic progression

Sal solves a very complicated algebraic trig problem that appeared as problem 29 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.

## Want to join the conversation?

• hey Sal..instead of replacing the sinA/a and SinC/c by SinB/b...try doing it by Replacing SinA/a by SinC/c and SinC/c by SinA/a...the 'a' and 'c' in the equation are cancelled out and we are left with
2(SinACosC + SinCCosA) = 2{Sin(A+B)}=2{Sin(180-B)}=2SinB
• It's really simple -

Using Sine Rule :
Sin A/a = Sin C/c
--> c/a = Sin C/Sin A
--> a/c = Sin A/Sin C
Substitute in Eq -
Sin A and Sin C cancel out leaving,
2(Sin A.Cos C+Sin C.Cos A)
It's in the form of [Sin (A+C)=Sin A.Cos C+Sin C.Cos A]
2Sin(A+C)
Then A+B+C=180
--> A+C=(180-B)
2Sin(180-B)
But Sin(180-B)=Sin B [Allied Angles]
2SinB
=2Sin60
=2.sqrt(3)/2
Ans=sqrt(3)
• Woah. At , when Sal started the drawing the right triangle, I had a moment of ferhoodle, and the cause of that was because my math teacher taught me that in a 30-60-90 triangle, the side opposite the 30 degree angle is 1, the hypotenuse is 2, and the only side left is √3. So when he said it's √3/2, I was a bit confused, especially since I learned this in a sophomore level geometry class, and this is a very reliable source. And when I searched up this predicament, all the finds on the Internet showed both sides of the spectrum, and that did nothing to aid me or even palliate my problem. So, in a nutshell, who is right and why? Thanks!!
• You're both right. You're thinking of it as a 1 - √3 - 2 triangle, and Sal is thinking of it as a 1/2 - √3/2 - 1 triangle. Those two triangles are similar, since your triangle's sides are all twice as large as Sal's, and both of those triangles have angles of 30, 60, and 90 degrees. You'll probably see the same thing with 45-45-90 triangles where some people will say that the sides are 1 - 1 - √2 and others will say that it is √2/2 - √2/2 - 1. Just depends on where you think the unit length should be. Hope this leaves you less ferhoodled!
• I am a tenther.
Can anyone help with this?

Square ABCD has length 13 and points E and F are exterior to the square such that BE=DF=5 and AE=CF=12.
FIND EF^2 ( that is , EF square)
• A simple way to solve this would be by extending AE and DF, let them intersect at G; extend EB and FC, let them intersect at H. Then EHFG is actually a square with side = 12 + 5 = 17. Thus the required length is the square of diagonal = 2 * 17^2 = 589 :D
• How is it that sin2C=2sinCcosC?
• sin(A+B)=sin(A).cos(B)+cos(A).sin(B)
sin(A+A)= sin(A).cos(A)+cos(A).sin(A)
sin(2A) =sin(A)[cos(A)+cos(A)]
sin(2A)= sin(A)[2cos(A)]
sin(2A)= 2sin(A).cos(A)
• Can anyone tell me where to find the rest of these problems? Sal starts at #29 but I'm interested in seeing the ones before this.
• You could do this much faster by concluding that if B=60 then A+C=120 (because A+B+C=180). Then using:

sin(A)/a=sin(C)/c

(multiply both sides by c and divide by sin(A))

c/a=sin(C)/sin(A)

also, taking the reciprocal

a/c=sin(A)/sin(C)

And there you go, now you just have to plug this into the beginning equation, that is

(a/c)sin(2C)+(c/a)sin(2A)

Replace a/c and c/a with what we found and use the fact that sin(2x)=2sin(x)cos(x)

[sin(A)/sin(C)]*2sin(C)cos(C)+[sin(C)/sin(A)]*2sin(A)cos(A)

Now cancel the sin(C)with the one in the denominator and also do the same thing for sin(A). Also factor out a 2 and you have

2[sin(A)cos(C)+sin(C)cos(A)]

Now use the fact that sin(x+y)=sin(x)cos(y)+sin(y)cos(x)

2[sin(A+C)]

We know that A+C=120 so sin(A+C)=sin(120)

2sin(120)

And also sin(120)=sin(180-60)=sin(60)=sqrt(3)/2 , and we know that the 2 in the denominator cancels with the one in our expresion so our answer in the end is sqrt(3)
• why is b (denominator) multiplied by b?? why not numerator sin B??
• sky hobo has explained so well...
(1 vote)
• Sir
can't we do it in this way ?
given a,b,care in arithmetic progression thus,
2B=A+C --(1)
but in any triangle,A+B+C=180 --(2)
SOLVING (1)&(2)
we get A=60 B=60 C=60
THUS THE TRIANGLE IS EQUILATERAL
therefore all sides are equal a=b=c
putting this values in the given eqn we get it's value = sqroot(3)