Math / Trigonometry

Question $B=75^{\circ}, c=4, a=3$ w window) 3 Points $\begin{array}{l} b=\square \\ C=\square \\ A=\square \end{array}$

Studdy Solution

STEP 1

What is this asking? We're given two sides of a triangle ( $a = 3$ and $c = 4$ ) and an angle ( $B = 75^\circ$ ).
We need to find the length of the third side ( $b$ ) and the measures of the other two angles ( $A$ and $C$ ). Watch out! Remember, the angles in a triangle always add up to $180^\circ$ !
Also, make sure your calculator is in degree mode, not radians!

STEP 2

1. Find side $b$
2. Find angle $C$
3. Find angle $A$

STEP 3

We're given two sides and the angle between them, so let's use the Law of Cosines to find the third side, $b$ .
The Law of Cosines says: $b^2 = a^2 + c^2 - 2ac \cdot \cos(B)$ .

STEP 4

Plug in our known values: $b^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos(75^\circ)$ .

STEP 5

Calculate $3^2 = 9$ and $4^2 = 16$ .
Also, $2 \cdot 3 \cdot 4 = 24$ .
So, $b^2 = 9 + 16 - 24 \cdot \cos(75^\circ)$ .

STEP 6

Now, $\cos(75^\circ) \approx 0.2588$ .
So, $b^2 \approx 25 - 24 \cdot 0.2588 \approx 25 - 6.2112 \approx 18.7888$ .

STEP 7

To find $b$ , we take the square root: $b \approx \sqrt{18.7888} \approx 4.33$ .
So, $b \approx 4.33$ .

STEP 8

Now that we have all three sides, we can use the Law of Sines to find angle $C$ .
The Law of Sines says: $\frac{\sin(C)}{c} = \frac{\sin(B)}{b}$ .

STEP 9

Plug in our values: $\frac{\sin(C)}{4} = \frac{\sin(75^\circ)}{4.33}$ .

STEP 10

Multiply both sides by $4$ to isolate $\sin(C)$ : $\sin(C) = 4 \cdot \frac{\sin(75^\circ)}{4.33} \approx 4 \cdot \frac{0.9659}{4.33} \approx 0.89$ .

STEP 11

To find $C$ , we take the inverse sine: $C = \arcsin(0.89) \approx 62.9^\circ$ .

STEP 12

We know that the sum of the angles in a triangle is $180^\circ$ .
So, $A + B + C = 180^\circ$ .

STEP 13

Plug in the values we know: $A + 75^\circ + 62.9^\circ = 180^\circ$ .

STEP 14

Combine the known angles: $A + 137.9^\circ = 180^\circ$ .

STEP 15

Subtract $137.9^\circ$ from both sides to find $A$ : $A = 180^\circ - 137.9^\circ = 42.1^\circ$ .

STEP 16

$b \approx 4.33$ , $C \approx 62.9^\circ$ , and $A \approx 42.1^\circ$ .

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Question $B=75^{\circ}, c=4, a=3$ w window) 3 Points $\begin{array}{l} b=\square \\ C=\square \\ A=\square \end{array}$

Studdy Solution

STEP 1

STEP 2

1. Find side $b$
2. Find angle $C$
3. Find angle $A$

STEP 3

We're given two sides and the angle between them, so let's use the Law of Cosines to find the third side, $b$ .
The Law of Cosines says: $b^2 = a^2 + c^2 - 2ac \cdot \cos(B)$ .

STEP 4

Plug in our known values: $b^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos(75^\circ)$ .

STEP 5

Calculate $3^2 = 9$ and $4^2 = 16$ .
Also, $2 \cdot 3 \cdot 4 = 24$ .
So, $b^2 = 9 + 16 - 24 \cdot \cos(75^\circ)$ .

STEP 6

Now, $\cos(75^\circ) \approx 0.2588$ .
So, $b^2 \approx 25 - 24 \cdot 0.2588 \approx 25 - 6.2112 \approx 18.7888$ .

STEP 7

To find $b$ , we take the square root: $b \approx \sqrt{18.7888} \approx 4.33$ .
So, $b \approx 4.33$ .

STEP 8

Now that we have all three sides, we can use the Law of Sines to find angle $C$ .
The Law of Sines says: $\frac{\sin(C)}{c} = \frac{\sin(B)}{b}$ .

STEP 9

Plug in our values: $\frac{\sin(C)}{4} = \frac{\sin(75^\circ)}{4.33}$ .

STEP 10

Multiply both sides by $4$ to isolate $\sin(C)$ : $\sin(C) = 4 \cdot \frac{\sin(75^\circ)}{4.33} \approx 4 \cdot \frac{0.9659}{4.33} \approx 0.89$ .

STEP 11

To find $C$ , we take the inverse sine: $C = \arcsin(0.89) \approx 62.9^\circ$ .

STEP 12

We know that the sum of the angles in a triangle is $180^\circ$ .
So, $A + B + C = 180^\circ$ .

STEP 13

Plug in the values we know: $A + 75^\circ + 62.9^\circ = 180^\circ$ .

STEP 14

Combine the known angles: $A + 137.9^\circ = 180^\circ$ .

STEP 15

Subtract $137.9^\circ$ from both sides to find $A$ : $A = 180^\circ - 137.9^\circ = 42.1^\circ$ .

STEP 16

$b \approx 4.33$ , $C \approx 62.9^\circ$ , and $A \approx 42.1^\circ$ .

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QuestionB=75∘,c=4,a=3B=75^{\circ}, c=4, a=3B=75∘,c=4,a=3 w window) 3 Points b=□C=□A=□\begin{array}{l} b=\square \\ C=\square \\ A=\square \end{array}b=□C=□A=□​

Studdy Solution

STEP 1

STEP 2

1. Find side bbb2. Find angle CCC3. Find angle AAA

STEP 3

We're given two sides and the angle between them, so let's **use the Law of Cosines** to find the third side, bbb.The Law of Cosines says: b2=a2+c2−2ac⋅cos⁡(B)b^2 = a^2 + c^2 - 2ac \cdot \cos(B)b2=a2+c2−2ac⋅cos(B).

STEP 4

**Plug in** our known values: b2=32+42−2⋅3⋅4⋅cos⁡(75∘)b^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos(75^\circ)b2=32+42−2⋅3⋅4⋅cos(75∘).

STEP 5

**Calculate** 32=93^2 = 932=9 and 42=164^2 = 1642=16.Also, 2⋅3⋅4=242 \cdot 3 \cdot 4 = 242⋅3⋅4=24.So, b2=9+16−24⋅cos⁡(75∘)b^2 = 9 + 16 - 24 \cdot \cos(75^\circ)b2=9+16−24⋅cos(75∘).

STEP 6

Now, cos⁡(75∘)≈0.2588\cos(75^\circ) \approx 0.2588cos(75∘)≈0.2588.So, b2≈25−24⋅0.2588≈25−6.2112≈18.7888b^2 \approx 25 - 24 \cdot 0.2588 \approx 25 - 6.2112 \approx 18.7888b2≈25−24⋅0.2588≈25−6.2112≈18.7888.

STEP 7

To find bbb, we take the **square root**: b≈18.7888≈4.33b \approx \sqrt{18.7888} \approx 4.33b≈18.7888​≈4.33.So, b≈4.33b \approx 4.33b≈4.33.

STEP 8

Now that we have all three sides, we can **use the Law of Sines** to find angle CCC.The Law of Sines says: sin⁡(C)c=sin⁡(B)b\frac{\sin(C)}{c} = \frac{\sin(B)}{b}csin(C)​=bsin(B)​.

STEP 9

**Plug in** our values: sin⁡(C)4=sin⁡(75∘)4.33\frac{\sin(C)}{4} = \frac{\sin(75^\circ)}{4.33}4sin(C)​=4.33sin(75∘)​.

STEP 10

**Multiply both sides by** 444 to isolate sin⁡(C)\sin(C)sin(C): sin⁡(C)=4⋅sin⁡(75∘)4.33≈4⋅0.96594.33≈0.89\sin(C) = 4 \cdot \frac{\sin(75^\circ)}{4.33} \approx 4 \cdot \frac{0.9659}{4.33} \approx 0.89sin(C)=4⋅4.33sin(75∘)​≈4⋅4.330.9659​≈0.89.

STEP 11

To find CCC, we take the **inverse sine**: C=arcsin⁡(0.89)≈62.9∘C = \arcsin(0.89) \approx 62.9^\circC=arcsin(0.89)≈62.9∘.

STEP 12

We know that the **sum of the angles in a triangle is** 180∘180^\circ180∘.So, A+B+C=180∘A + B + C = 180^\circA+B+C=180∘.

STEP 13

**Plug in** the values we know: A+75∘+62.9∘=180∘A + 75^\circ + 62.9^\circ = 180^\circA+75∘+62.9∘=180∘.

STEP 14

**Combine** the known angles: A+137.9∘=180∘A + 137.9^\circ = 180^\circA+137.9∘=180∘.

STEP 15

**Subtract** 137.9∘137.9^\circ137.9∘ from both sides to find AAA: A=180∘−137.9∘=42.1∘A = 180^\circ - 137.9^\circ = 42.1^\circA=180∘−137.9∘=42.1∘.

STEP 16

b≈4.33b \approx 4.33b≈4.33, C≈62.9∘C \approx 62.9^\circC≈62.9∘, and A≈42.1∘A \approx 42.1^\circA≈42.1∘.

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Question $B=75^{\circ}, c=4, a=3$ w window) 3 Points $\begin{array}{l} b=\square \\ C=\square \\ A=\square \end{array}$

1. Find side $b$
2. Find angle $C$
3. Find angle $A$

We're given two sides and the angle between them, so let's use the Law of Cosines to find the third side, $b$ .
The Law of Cosines says: $b^2 = a^2 + c^2 - 2ac \cdot \cos(B)$ .

Plug in our known values: $b^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos(75^\circ)$ .

Calculate $3^2 = 9$ and $4^2 = 16$ .
Also, $2 \cdot 3 \cdot 4 = 24$ .
So, $b^2 = 9 + 16 - 24 \cdot \cos(75^\circ)$ .

Now, $\cos(75^\circ) \approx 0.2588$ .
So, $b^2 \approx 25 - 24 \cdot 0.2588 \approx 25 - 6.2112 \approx 18.7888$ .

To find $b$ , we take the square root: $b \approx \sqrt{18.7888} \approx 4.33$ .
So, $b \approx 4.33$ .

Now that we have all three sides, we can use the Law of Sines to find angle $C$ .
The Law of Sines says: $\frac{\sin(C)}{c} = \frac{\sin(B)}{b}$ .

Plug in our values: $\frac{\sin(C)}{4} = \frac{\sin(75^\circ)}{4.33}$ .

Multiply both sides by $4$ to isolate $\sin(C)$ : $\sin(C) = 4 \cdot \frac{\sin(75^\circ)}{4.33} \approx 4 \cdot \frac{0.9659}{4.33} \approx 0.89$ .

To find $C$ , we take the inverse sine: $C = \arcsin(0.89) \approx 62.9^\circ$ .

We know that the sum of the angles in a triangle is $180^\circ$ .
So, $A + B + C = 180^\circ$ .

Plug in the values we know: $A + 75^\circ + 62.9^\circ = 180^\circ$ .

Combine the known angles: $A + 137.9^\circ = 180^\circ$ .

Subtract $137.9^\circ$ from both sides to find $A$ : $A = 180^\circ - 137.9^\circ = 42.1^\circ$ .

$b \approx 4.33$ , $C \approx 62.9^\circ$ , and $A \approx 42.1^\circ$ .