SOLUTION: cos alpha= 1/2and tan beta=1/3.find sin(alpha+beta),where alpha and beta are both acute angles

Question 901286: cos alpha= 1/2and tan beta=1/3.find sin(alpha+beta),where alpha and beta are both acute angles

Found 2 solutions by Theo, harpazo:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you have cosine(a) = 1/2
this means that adjacent side is 1 and hypotenuse is 2.
use pythagorus to find that opposite side is sqrt(3)

you have tan(b) = 1/3
this means that opposite side is 1 and adjacent side is 3.
use pythagorus to find that hypotenuse is sqrt(10)

pythagorus is a^2 + b^2 = c^2 where c is the hypotenuse and a and b are legs.

sin(a+b) = sin(a) * cos(b) + cos(a) * sin(b)

sin(a) = sqrt(3)/2
cos(b) = 3/sqrt(10)
cos(a) = 1/sqrt(10)
sin(b) = 1/2

formula becomes:

sin(a+b) = sqrt(3)/2 * 3/sqrt(10) + 1/sqrt(10) * 1/2

cimplify that to get:

(sqrt(3)*3)/(2*sqrt(10) + (1*1)/(2*sqrt(10)

combine into one common denominator and you get:

(3*sqrt(3) + (1*1) / (2*sqrt(10)

use your calculator to solve to get sin(a+b) = .9796927193

take the arcsine of that to get (a+b) = 78.43494882 degrees.

a = 60 degrees.
b = 18.43494882
a+b = 78.43494882 degrees.

you were asked to find sin(a+b), so your solution is:

sin(a+b) = .9796927193

Answer by harpazo(655) (Show Source): You can put this solution on YOUR website!
We need to find sin alpha and sin beta.
cosine = adjacent over hypotenuse.
x^2 + 1^2 = 2^2
x^2 = 4 - 1
x^2 = 3
sqrt{x^2} = sqrt{3)
x = sqrt{3)
sin alpha = sqrt{3}/2
tangent = opposite over adjacent.
1^2 + 3^2 = x^2
1 + 9 = x^2
10 = x^2
sqrt{10} = sqrt{x^2}
sqrt{10} = x
sin beta = 1/sqrt{10}
We now plug and chug.
sin(A + B) = sin A cos B + cos A sin B
Let A = alpha and B = beta for short.
We have sin alpha and sin beta.
We need cos beta and cos alpha.
cos alpha = 1/2 and cos beta = 3/sqrt{10}.
Plug into formula and simplify.
Here is the formula you need:
sin(A + B) = sin A times cos B + cos A times sin B
Can you finish?

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