Find tan(B + C) given that sin C = 1/4 , with C in quadrant II, and sin B= -1/2 with B in quadrant 4?

Question

a) 1-3√5/8
b) 4√3+√15/-11
c) 4√3 - √15/11
d) 1+3√5/8

Answer 1

Hi,

We can solve this problem using this:

tan(B + C) = tanB + tanC / 1 - (tanB)(tanC)

so, let's start with B.

if sin(B) = (-1/2), and is in the IV quadrant, so
sin(3pi/2 + x) = -1/2, where (3pi/2 + x) is B.
(sin3pi/2)(cosx) + (cos3pi/2)(sinx) = -1/2
then,
(-1)(cosx) + (0)(sinx) = -1/2
-cosx = -1/2
cosx = 1/2
x= 60 degrees
and B is 330 degrees.
therefore, tanB = -1/sqrt(3)
now let's work with C.

sin(C) = 1/4
so doing a right triangle we have that:

(1)^2 + (x)^2 = 4^2
x^2 = 15
so, x is equal to: sqrt(15), thus, we know that value of
tan(C) = (-)1 / sqrt15, it's negative because C is on the second quadrant

Therefore,

tan(B + C) = tanB + tanC / 1 - (tanB)(tanC)
tan(B + C) = (-1/sqrt3) + (-1/sqrt15) / 1 - (-1/sqrt3) (-1/sqrt15)
tan(B + C)= (-3)(sqrt(5) + 1) / 3sqrt(15) - sqrt(3)
tan(B+ C) = -9.708203933 / 9.886899231
tan(B + C) = -0.981926052

I hope it help

Answer 2

sin C=1/4
C=arcsin 1/4=14.478º in the 2nd quadrant, C=165.522º
Sin B=-1/2
B=arcsin -1/2=330º
B+C=330º+165.522º=495.5224878140700761212289652
tan(B+C)=-0.98192605240753873266264097870963

Answer 3

If you know Sin of an angle, you can compute Cos of the angle from Cos^2 @ = 1- Sin^2 @. Then you can evaluate sin @ / cos @, being aware that in the second quadrant, tan @ is negative. That should get you moving towards an answer in radical form.

Answer 4

given sin(A+B-C) = a million/2= sin30 so A+B-C=30 A+B= 30+C and in a triagle A+B+C=ninety B+C= ninety-A Cos(B+C) = cos (ninety-A) =SinA now evaluate cos (B+C-A) enable B+C=D cos(D-A) =cosCcosD- sinCsinD = cos(B+A)cosA- sin(B+C)sinA

Answer 5

sin(C) = 1/4

1^2 + b^2 = 4^2
1 + b^2 = 16
b^2 = 15
b = sqrt(15)

cos(C) = -sqrt(15)/4
tan(C) = -1/sqrt(15)

sin(B) = -1/2

(-1)^2 + b^2 = 2^2
1 + b^2 = 4
b^2 = 3
b = sqrt(3)

cos(B) = sqrt(3)/2
cos(B) = -1/sqrt(3)

tan(B + C) = (tan(B) + tan(C)) / (1 - tan(B)tan(C))

tan(B + C) = ((-1/sqrt(15)) + (-1/sqrt(3)))/(1 - ((-1/sqrt(15))(-1/sqrt(3)))

tan(B + C) = ((-sqrt(3) - sqrt(15))/sqrt(45))/(1 - (1/sqrt(45)))

tan(B + C) = ((-sqrt(3) - sqrt(15))/sqrt(45))/((sqrt(45) - 1)/sqrt(45))

tan(B + C) = ((-sqrt(3) - sqrt(15))/sqrt(45)) * (sqrt(45)/(sqrt(45) - 1))

tan(B + C) = (-sqrt(3) - sqrt(15))/(sqrt(45) - 1)