Math / Geometry

QuestionIn the following exercise, two sides and an angle are given. First determine whether the information results in no triangle, one triangle, or two triangles. Solve the resulting triangle. $\mathrm{a}=9.2, \mathrm{~b}=7.4 \text {, and } \mathrm{A}=38^{\circ}$ B. There are two triangles. The angle corresponding to the triangle containing $B_{1}$ is $C_{1} \approx$ $\square$ ${ }^{\circ}$ . The angle corresponding to the triangle containing $\mathrm{B}_{2}$ is $\mathrm{C}_{2} \approx$ $\square$ $\because$ (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 $c \approx$ $\square$ - (Round to one decimal place as needed.) B. There are two triangles. The corresponding length of side c for each triangle is $\mathrm{c}_{1} \approx$ $\square$ and $\mathrm{c}_{2} \approx$ $\square$ . (Round to one decimal place as needed) C. There is no solution.

Studdy Solution

STEP 1

What is this asking? Given two sides of a triangle (a = 9.2 and b = 7.4) and an angle (A = 38°), we need to figure out if we can make a triangle, and if so, how many different triangles can we make, and what are the missing sides and angles? Watch out! Remember the ambiguous case of the sine rule!
Sometimes, with this setup, you can get two possible triangles, one acute and one obtuse.

STEP 2

1. Check for possible triangles
2. Calculate angle B
3. Calculate angle C and side c for the first triangle
4. Check for a second triangle
5. Calculate angle C and side c for the second triangle (if it exists)

STEP 3

Let's compare $a \cdot \sin(A)$ with $b$ .
If $a \cdot \sin(A) < b < a$ , then there are two possible triangles.
If $b \sin(A) = a$ , there is one triangle.
If $b < a \sin(A)$ , there is no triangle.
If $b > a$ , there is one triangle.

STEP 4

Let's calculate $a \cdot \sin(A)$ : $9.2 \cdot \sin(38^\circ) \approx 9.2 \cdot 0.6157 \approx 5.664$

STEP 5

Since $5.664 < 7.4 < 9.2$ , we have two possible triangles!
Get ready for some exciting triangle-solving action!

STEP 6

We can use the sine rule to find angle B: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$ $\frac{\sin(38^\circ)}{9.2} = \frac{\sin(B)}{7.4}$ $\sin(B) = \frac{7.4 \cdot \sin(38^\circ)}{9.2} \approx \frac{7.4 \cdot 0.6157}{9.2} \approx 0.4945$

STEP 7

Now, we find the inverse sine: $B_1 = \arcsin(0.4945) \approx 29.6^\circ$ This is our first angle B!

STEP 8

Since the angles in a triangle add up to 180°, we can find $C_1$ : $C_1 = 180^\circ - A - B_1 \approx 180^\circ - 38^\circ - 29.6^\circ \approx 112.4^\circ$

STEP 9

Now, let's use the sine rule again to find $c_1$ : $\frac{\sin(A)}{a} = \frac{\sin(C_1)}{c_1}$ $c_1 = \frac{a \cdot \sin(C_1)}{\sin(A)} \approx \frac{9.2 \cdot \sin(112.4^\circ)}{\sin(38^\circ)} \approx \frac{9.2 \cdot 0.9249}{0.6157} \approx 13.8$

STEP 10

Remember, sine is positive in both the first and second quadrants.
So, there might be another angle B that works!
Let's find it: $B_2 = 180^\circ - B_1 \approx 180^\circ - 29.6^\circ \approx 150.4^\circ$

STEP 11

Let's check if $A + B_2 < 180^\circ$ : $38^\circ + 150.4^\circ = 188.4^\circ$ Since this is greater than $180^\circ$ , there is no second triangle.
Our initial calculation for $B_1$ was the only valid solution.

STEP 12

Since there is no second triangle, we can skip this step!

STEP 13

There is only one triangle.
The approximate values are: $B \approx 29.6^\circ$ , $C \approx 112.4^\circ$ , and $c \approx 13.8$ .
So we choose option A, and fill in 13.8.

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Studdy Solution

STEP 1

STEP 2

1. Check for possible triangles
2. Calculate angle B
3. Calculate angle C and side c for the first triangle
4. Check for a second triangle
5. Calculate angle C and side c for the second triangle (if it exists)

STEP 3

Let's compare $a \cdot \sin(A)$ with $b$ .
If $a \cdot \sin(A) < b < a$ , then there are two possible triangles.
If $b \sin(A) = a$ , there is one triangle.
If $b < a \sin(A)$ , there is no triangle.
If $b > a$ , there is one triangle.

STEP 4

Let's calculate $a \cdot \sin(A)$ : $9.2 \cdot \sin(38^\circ) \approx 9.2 \cdot 0.6157 \approx 5.664$

STEP 5

Since $5.664 < 7.4 < 9.2$ , we have two possible triangles!
Get ready for some exciting triangle-solving action!

STEP 6

We can use the sine rule to find angle B: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$ $\frac{\sin(38^\circ)}{9.2} = \frac{\sin(B)}{7.4}$ $\sin(B) = \frac{7.4 \cdot \sin(38^\circ)}{9.2} \approx \frac{7.4 \cdot 0.6157}{9.2} \approx 0.4945$

STEP 7

Now, we find the inverse sine: $B_1 = \arcsin(0.4945) \approx 29.6^\circ$ This is our first angle B!

STEP 8

Since the angles in a triangle add up to 180°, we can find $C_1$ : $C_1 = 180^\circ - A - B_1 \approx 180^\circ - 38^\circ - 29.6^\circ \approx 112.4^\circ$

STEP 9

Now, let's use the sine rule again to find $c_1$ : $\frac{\sin(A)}{a} = \frac{\sin(C_1)}{c_1}$ $c_1 = \frac{a \cdot \sin(C_1)}{\sin(A)} \approx \frac{9.2 \cdot \sin(112.4^\circ)}{\sin(38^\circ)} \approx \frac{9.2 \cdot 0.9249}{0.6157} \approx 13.8$

STEP 10

Remember, sine is positive in both the first and second quadrants.
So, there might be another angle B that works!
Let's find it: $B_2 = 180^\circ - B_1 \approx 180^\circ - 29.6^\circ \approx 150.4^\circ$

STEP 11

Let's check if $A + B_2 < 180^\circ$ : $38^\circ + 150.4^\circ = 188.4^\circ$ Since this is greater than $180^\circ$ , there is no second triangle.
Our initial calculation for $B_1$ was the only valid solution.

STEP 12

Since there is no second triangle, we can skip this step!

STEP 13

There is only one triangle.
The approximate values are: $B \approx 29.6^\circ$ , $C \approx 112.4^\circ$ , and $c \approx 13.8$ .
So we choose option A, and fill in 13.8.

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Studdy Solution

STEP 1

STEP 2

1. Check for possible triangles2. Calculate angle B3. Calculate angle C and side c for the first triangle4. Check for a second triangle5. Calculate angle C and side c for the second triangle (if it exists)

STEP 3

STEP 4

Let's **calculate** a⋅sin⁡(A)a \cdot \sin(A)a⋅sin(A): 9.2⋅sin⁡(38∘)≈9.2⋅0.6157≈5.6649.2 \cdot \sin(38^\circ) \approx 9.2 \cdot 0.6157 \approx 5.6649.2⋅sin(38∘)≈9.2⋅0.6157≈5.664

STEP 5

Since 5.664<7.4<9.25.664 < 7.4 < 9.25.664<7.4<9.2, we have **two possible triangles**!Get ready for some exciting triangle-solving action!

STEP 6

STEP 7

Now, we find the **inverse sine**: B1=arcsin⁡(0.4945)≈29.6∘B_1 = \arcsin(0.4945) \approx 29.6^\circB1​=arcsin(0.4945)≈29.6∘ This is our first angle B!

STEP 8

Since the **angles in a triangle add up to 180°**, we can find C1C_1C1​: C1=180∘−A−B1≈180∘−38∘−29.6∘≈112.4∘C_1 = 180^\circ - A - B_1 \approx 180^\circ - 38^\circ - 29.6^\circ \approx 112.4^\circC1​=180∘−A−B1​≈180∘−38∘−29.6∘≈112.4∘

STEP 9

STEP 10

Remember, sine is positive in both the first and second quadrants.So, there might be another angle B that works!Let's find it: B2=180∘−B1≈180∘−29.6∘≈150.4∘B_2 = 180^\circ - B_1 \approx 180^\circ - 29.6^\circ \approx 150.4^\circB2​=180∘−B1​≈180∘−29.6∘≈150.4∘

STEP 11

Let's check if A+B2<180∘A + B_2 < 180^\circA+B2​<180∘: 38∘+150.4∘=188.4∘38^\circ + 150.4^\circ = 188.4^\circ38∘+150.4∘=188.4∘ Since this is greater than 180∘180^\circ180∘, there is no second triangle.Our initial calculation for B1B_1B1​ was the only valid solution.

STEP 12

Since there is no second triangle, we can skip this step!

STEP 13

There is only one triangle.The approximate values are: B≈29.6∘B \approx 29.6^\circB≈29.6∘, C≈112.4∘C \approx 112.4^\circC≈112.4∘, and c≈13.8c \approx 13.8c≈13.8.So we choose option A, and fill in 13.8.

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1. Check for possible triangles
2. Calculate angle B
3. Calculate angle C and side c for the first triangle
4. Check for a second triangle
5. Calculate angle C and side c for the second triangle (if it exists)

Let's calculate $a \cdot \sin(A)$ : $9.2 \cdot \sin(38^\circ) \approx 9.2 \cdot 0.6157 \approx 5.664$

Since $5.664 < 7.4 < 9.2$ , we have two possible triangles!
Get ready for some exciting triangle-solving action!

Now, we find the inverse sine: $B_1 = \arcsin(0.4945) \approx 29.6^\circ$ This is our first angle B!

Since the angles in a triangle add up to 180°, we can find $C_1$ : $C_1 = 180^\circ - A - B_1 \approx 180^\circ - 38^\circ - 29.6^\circ \approx 112.4^\circ$

Remember, sine is positive in both the first and second quadrants.
So, there might be another angle B that works!
Let's find it: $B_2 = 180^\circ - B_1 \approx 180^\circ - 29.6^\circ \approx 150.4^\circ$

Let's check if $A + B_2 < 180^\circ$ : $38^\circ + 150.4^\circ = 188.4^\circ$ Since this is greater than $180^\circ$ , there is no second triangle.
Our initial calculation for $B_1$ was the only valid solution.

There is only one triangle.
The approximate values are: $B \approx 29.6^\circ$ , $C \approx 112.4^\circ$ , and $c \approx 13.8$ .
So we choose option A, and fill in 13.8.