Math  /  Geometry

QuestionIn the following exercise, two sides and an angle are given. First determine whether the information results in no triangle, one tr a=9.9, b=7.6\mathrm{a}=9.9, \mathrm{~b}=7.6, and A=37\mathrm{A}=37^{\circ} \qquad (Round to one decimal place as needed.) C. There is no solution.
Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is only one triangle where C\mathrm{C} \approx \square { }^{\circ}. (Round to ofie decimal place as needed.) B. There are two triangles. The angle corresponding to the triangle containing B1\mathrm{B}_{1} is C1\mathrm{C}_{1} \approx \square - The angle corresponding to the (Round to one decimal place as needed.) C. There is no solution.

Studdy Solution

STEP 1

What is this asking? We're given two sides of a triangle (a=9.9a = 9.9 and b=7.6b = 7.6) and an angle (A=37A = 37^\circ), and we need to figure out if we can make a triangle with these measurements and, if so, find the missing angle CC. Watch out! Remember, not just any three measurements can form a triangle!
Sometimes, the given sides and angles just won't fit together.
Also, be super careful with your calculations, especially when using sine, cosine, and their inverses.

STEP 2

1. Law of Sines to find angle B
2. Check for ambiguous case
3. Calculate angle C

STEP 3

We're given aa, bb, and AA, and we want to find BB.
The **Law of Sines** connects these: sinAa=sinBb\frac{\sin A}{a} = \frac{\sin B}{b}.
Let's plug in what we know: sin379.9=sinB7.6\frac{\sin 37^\circ}{9.9} = \frac{\sin B}{7.6}.

STEP 4

To get sinB\sin B by itself, we'll multiply both sides of the equation by 7.67.6: sinB=7.6sin379.9\sin B = \frac{7.6 \cdot \sin 37^\circ}{9.9}.

STEP 5

Now, let's crunch the numbers: sinB7.60.60189.94.57379.90.462\sin B \approx \frac{7.6 \cdot 0.6018}{9.9} \approx \frac{4.5737}{9.9} \approx 0.462.

STEP 6

To find BB, we'll take the inverse sine (arcsin) of both sides: B=arcsin(0.462)B = \arcsin(0.462).
This gives us B27.5B \approx 27.5^\circ.

STEP 7

Since we used the Law of Sines and found an angle using inverse sine, there's a *chance* we have the ambiguous case, meaning there might be *two* possible triangles.
Let's investigate!

STEP 8

The other possible angle for BB is the supplement of the first one: B2=18027.5=152.5B_2 = 180^\circ - 27.5^\circ = 152.5^\circ.

STEP 9

Now, let's add this to our known angle AA: A+B2=37+152.5=189.5A + B_2 = 37^\circ + 152.5^\circ = 189.5^\circ.
Uh oh!
This is already bigger than 180180^\circ, and triangles can only have 180180^\circ total, so B2B_2 can't be a valid angle for our triangle.
That means we only have *one* possible triangle!

STEP 10

We know that the angles in a triangle add up to 180180^\circ, so A+B+C=180A + B + C = 180^\circ.

STEP 11

We know A=37A = 37^\circ and B27.5B \approx 27.5^\circ, so 37+27.5+C=18037^\circ + 27.5^\circ + C = 180^\circ.
This simplifies to 64.5+C=18064.5^\circ + C = 180^\circ.
Subtracting 64.564.5^\circ from both sides gives us C18064.5=115.5C \approx 180^\circ - 64.5^\circ = 115.5^\circ.

STEP 12

There is only one possible triangle, and the angle CC is approximately 115.5115.5^\circ.

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