# I cannot figure these two problems out. Thank you for you help, I really appreciate it! 5. Solve the triangle (see the image.) 6. Calculate the exact value of sin(x + y) if sin(x) = 1/3 and angle...

I cannot figure these two problems out. Thank you for you help, I really appreciate it!

5. Solve the triangle (see the image.)

6. Calculate the exact value of sin(x + y) if sin(x) = 1/3 and angle x ends in the second quadrant and cos(y) = 1/5 and angle y is in the first quadrant.

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5. Since the angles `alpha ` and `beta ` are given, the third angle of the triangle can be found using the fact that the sum of angles in a triangle is 18 degrees:

`alpha + beta + gamma = 180 `

`41+ 33 + gamma=180 `

`gamma =180-33-41=106 `

Now, since the side c is given, we can use the Sine theorem to find remaining sides. The Sine theorem says that in a triangle all ratios of sides to the sines of opposite angles are equal:

`a/sin(alpha) = b/sin(beta) = c/sin(gamma) `

`c/sin(gamma) = 21.85 `

From here, we can solve for a and b:

`a=21.85*sin(alpha) =14.33 `

`b=21.85*sin(beta) =11.9 `

**So, the unknown quantities in the given triangle are**

**`gamma=106 ` degrees**

**a = 14.33, b = 11.9**

6. We need to use the Sine of Sum formula to solve this problem:

sin(x + y) = sin(x)cos(y) + cos(x)sin(y)

We already know sin(x) and cos(y), but we don't know cos(x) and sin(y). They can be found using Pythagorean theorem:

`sin^2(x) + cos^2(x) = 1 `

`cos(x) = +-sqrt(1-sin^2(x)) = +-sqrt(1-1/9) = +-sqrt(8)/3=+-(2sqrt(2))/3 `

Since angle x is in the second quadrant, its cosine is negative, so

` cos(x) = -(2sqrt2)/3`

Similarly, for angle y

`sin^2(y) + cos^2(y) = 1 `

`sin(y) = +-sqrt(1-cos^2(y)) = +-sqrt(1-1/25)) = +-sqrt(24)/5 = +-(2sqrt6)/5 `

Since angle y is in the first quadrant, its sine is positive.

`sin(y) = (2sqrt(6))/5 `

Plug in these values into the formula:

`sin(x + y) = 1/3 * 1/5 + (-2sqrt(2))/3 * (2sqrt(6))/5 = 1/15 - (4sqrt(12))/15=(1- 8sqrt(3))/15 `

**The exact answer for sin(x + y) is ` (1-8sqrt(3))/15` .**

6. x-ends in the second quadrant. sine function is positive here, and cosine - negative. y-ends in the first quadrant, where both sine and cosine functions are positive. Again, (x+y) must end in the third quadrant where sine function is negative.

`sinx=1/3`

Applying the trigonometric identity `sin^2theta+cos^2theta=1 `

`cosx=sqrt(1-(1/3)^2)`

`=-(2sqrt2)/3` .

Again, `cosy=1/5`

Similarly,

`siny=sqrt(1-(1/5)^2)`

`=(2sqrt6)/5`

From the formula of sum of angles,

`sin(x+y)=sinxcosy+cosxsiny`

`=1/3xx1/5+(-(2sqrt2)/3)xx(2sqrt6)/5`

`=1/15-(8sqrt3)/15`

`=(1-8sqrt3)/15`

As expected, this function has a negative value.