Math / Geometry

QuestionSolve the triangle. Round all angles to the nearest degree. $A \approx 46{ }^{\circ}$ (Do not round until the final answer. Then round to the nearest degree as needed.) B $\approx$ $\square$ - (Do not round until the final answer. Then round to the nearest degree as needed.) $\mathrm{C} \approx$ $\square$ $]^{\circ}$ (Do not round until the final answer. Then round to the nearest degree as needed.)

Studdy Solution

STEP 1

What is this asking? We're given a triangle with two sides equal ( $b$ and $c$ ), and we know one angle ( $A$ ) and the lengths of all three sides.
We need to find the other two angles ( $B$ and $C$ ). Watch out! Remember, since two sides are equal ( $b=c$ ), the angles opposite those sides ( $B$ and $C$ ) will also be equal!
Don't get tricked into thinking you have to do twice the work!

STEP 2

1. Find angle $B$ using the Law of Sines.
2. Find angle $C$ .

STEP 3

The Law of Sines says that the ratio of the sine of an angle to the length of its opposite side is the same for all angles and sides in a triangle.
So we can write: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$ We know $A \approx 46^\circ$ , $a = 7$ , and $b = 9$ , so let's plug those values in!

STEP 4

$\frac{\sin(46^\circ)}{7} = \frac{\sin(B)}{9}$ Now, we want to isolate $\sin(B)$ , so we'll multiply both sides of the equation by 9.

STEP 5

$\sin(B) = \frac{9 \cdot \sin(46^\circ)}{7}$

STEP 6

$\sin(B) \approx \frac{9 \cdot \textbf{0.7193}}{7} \approx \frac{\textbf{6.4737}}{7} \approx \textbf{0.9248}$

STEP 7

To find $B$ , we take the inverse sine (arcsin) of 0.9248: $B = \arcsin(\textbf{0.9248}) \approx \textbf{67.65}^\circ$

STEP 8

Rounding to the nearest degree, we get $B \approx \textbf{68}^\circ$ .

STEP 9

Since sides $b$ and $c$ are equal ( $b=c=9$ ), angles $B$ and $C$ must also be equal!

STEP 10

Therefore, $C \approx B \approx \textbf{68}^\circ$ .

STEP 11

$B \approx 68^\circ$ $C \approx 68^\circ$

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Studdy Solution

STEP 1

STEP 2

1. Find angle $B$ using the Law of Sines.
2. Find angle $C$ .

STEP 3

The Law of Sines says that the ratio of the sine of an angle to the length of its opposite side is the same for all angles and sides in a triangle.
So we can write: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$ We know $A \approx 46^\circ$ , $a = 7$ , and $b = 9$ , so let's plug those values in!

STEP 4

$\frac{\sin(46^\circ)}{7} = \frac{\sin(B)}{9}$ Now, we want to isolate $\sin(B)$ , so we'll multiply both sides of the equation by 9.

STEP 5

$\sin(B) = \frac{9 \cdot \sin(46^\circ)}{7}$

STEP 6

$\sin(B) \approx \frac{9 \cdot \textbf{0.7193}}{7} \approx \frac{\textbf{6.4737}}{7} \approx \textbf{0.9248}$

STEP 7

To find $B$ , we take the inverse sine (arcsin) of 0.9248: $B = \arcsin(\textbf{0.9248}) \approx \textbf{67.65}^\circ$

STEP 8

Rounding to the nearest degree, we get $B \approx \textbf{68}^\circ$ .

STEP 9

Since sides $b$ and $c$ are equal ( $b=c=9$ ), angles $B$ and $C$ must also be equal!

STEP 10

Therefore, $C \approx B \approx \textbf{68}^\circ$ .

STEP 11

$B \approx 68^\circ$ $C \approx 68^\circ$

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Studdy Solution

STEP 1

STEP 2

1. Find angle BBB using the Law of Sines.2. Find angle CCC.

STEP 3

STEP 4

sin⁡(46∘)7=sin⁡(B)9 \frac{\sin(46^\circ)}{7} = \frac{\sin(B)}{9} 7sin(46∘)​=9sin(B)​ Now, we want to isolate sin⁡(B)\sin(B)sin(B), so we'll multiply both sides of the equation by **9**.

STEP 5

sin⁡(B)=9⋅sin⁡(46∘)7 \sin(B) = \frac{9 \cdot \sin(46^\circ)}{7} sin(B)=79⋅sin(46∘)​

STEP 6

sin⁡(B)≈9⋅0.71937≈6.47377≈0.9248 \sin(B) \approx \frac{9 \cdot \textbf{0.7193}}{7} \approx \frac{\textbf{6.4737}}{7} \approx \textbf{0.9248} sin(B)≈79⋅0.7193​≈76.4737​≈0.9248

STEP 7

To find BBB, we take the inverse sine (arcsin) of **0.9248**: B=arcsin⁡(0.9248)≈67.65∘ B = \arcsin(\textbf{0.9248}) \approx \textbf{67.65}^\circ B=arcsin(0.9248)≈67.65∘

STEP 8

Rounding to the nearest degree, we get B≈68∘B \approx \textbf{68}^\circB≈68∘.

STEP 9

Since sides bbb and ccc are equal (b=c=9b=c=9b=c=9), angles BBB and CCC must also be equal!

STEP 10

Therefore, C≈B≈68∘C \approx B \approx \textbf{68}^\circC≈B≈68∘.

STEP 11

B≈68∘B \approx 68^\circB≈68∘ C≈68∘C \approx 68^\circC≈68∘

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1. Find angle $B$ using the Law of Sines.
2. Find angle $C$ .

$\frac{\sin(46^\circ)}{7} = \frac{\sin(B)}{9}$ Now, we want to isolate $\sin(B)$ , so we'll multiply both sides of the equation by 9.

$\sin(B) = \frac{9 \cdot \sin(46^\circ)}{7}$

$\sin(B) \approx \frac{9 \cdot \textbf{0.7193}}{7} \approx \frac{\textbf{6.4737}}{7} \approx \textbf{0.9248}$

To find $B$ , we take the inverse sine (arcsin) of 0.9248: $B = \arcsin(\textbf{0.9248}) \approx \textbf{67.65}^\circ$

Rounding to the nearest degree, we get $B \approx \textbf{68}^\circ$ .

Since sides $b$ and $c$ are equal ( $b=c=9$ ), angles $B$ and $C$ must also be equal!

Therefore, $C \approx B \approx \textbf{68}^\circ$ .

$B \approx 68^\circ$ $C \approx 68^\circ$