Question

1. a. determine side a .

b. determine ∠b .
c. determine ∠c .

33.in a scalene triangle the measurement of the 3 angles are 37.60 degrees b and c.
the measurements of the corresponding opposite sides are, a, 48.000 millimeters and
50.900 millimeters.

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that in any triangle, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, where $a, b, c$ are the lengths of the sides opposite angles $A, B, C$ respectively.

From the problem, we know one angle is $37.60^\circ$ (let's assume angle $A = 37.60^\circ$), side $a$ is opposite angle $A$, side $b = 50.900$ mm (opposite angle $B$), and side $c = 48.000$ mm (opposite angle $C$)? Wait, no, re - reading the details: "the measurement of the 3 angles are 37.60 degrees B and C" – maybe a typo, probably one angle is $37.60^\circ$, let's assume angle $A = 37.60^\circ$, side $a$ is unknown, side $b = 50.900$ mm (opposite angle $B$), side $c = 48.000$ mm (opposite angle $C$). Wait, no, the diagram: the side of length $48.000$ mm, side of length $50.900$ mm, and side $a$. Let's correct: Let's say angle $A = 37.60^\circ$, side $a$ is opposite angle $A$, side $b = 50.900$ mm (opposite angle $B$), side $c = 48.000$ mm (opposite angle $C$).

Step1 (for part a: Determine side a)

Using the Law of Sines: $\frac{a}{\sin A}=\frac{c}{\sin C}$? Wait, no, maybe we first find angle $B$ or $C$? Wait, the triangle is scalene, so all angles and sides are different. Wait, the sum of angles in a triangle is $180^\circ$. Let's assume that we have two sides and a non - included angle? Wait, no, let's re - examine the problem statement. The details say: "In a scalene triangle the measurement of the 3 angles are 37.60 degrees B and C. The measurements of the corresponding opposite sides are, a, 48.000 millimeters and 50.900 millimeters." So, angle $A$ is not given? Wait, maybe a misstatement. Let's assume that angle $A = 37.60^\circ$, side $a$ is opposite angle $A$, side $b = 50.900$ mm (opposite angle $B$), side $c = 48.000$ mm (opposite angle $C$).

Wait, maybe the correct approach is: Let's use the Law of Sines. Let's suppose that we know two sides and we need to find the third side or angles.

Part a: Determine side a

We know two sides: let's say side $c = 48.000$ mm, side $b = 50.900$ mm, and angle $A = 37.60^\circ$. Wait, no, the Law of Cosines: $a^{2}=b^{2}+c^{2}-2bc\cos A$. Let's try that. If angle $A = 37.60^\circ$, $b = 50.900$ mm, $c = 48.000$ mm.

Then $a^{2}=(50.900)^{2}+(48.000)^{2}-2\times50.900\times48.000\times\cos(37.60^\circ)$

First, calculate $(50.900)^{2}=50.9^{2}=2590.81$, $(48.000)^{2}=2304$

$\cos(37.60^\circ)\approx\cos(37.6)\approx0.791$

Then $2\times50.9\times48 = 2\times2443.2 = 4886.4$

$4886.4\times0.791\approx4886.4\times0.791\approx3865.14$

Then $a^{2}=2590.81 + 2304-3865.14=2590.81 + 2304=4894.81-3865.14 = 1029.67$

$a=\sqrt{1029.67}\approx32.09$ mm? Wait, that seems small. Maybe my assumption of angle $A$ is wrong.

Wait, maybe the angle of $37.60^\circ$ is angle $C$, side $c = 48.000$ mm, side $b = 50.900$ mm, and we need to find side $a$ and angles $B$ and $A$.

Let's start over.

Let's define:

- Let side $a$ be the side we need to find.

- Side $b = 50.900$ mm (opposite angle $B$)

- Side $c = 48.000$ mm (opposite angle $C = 37.60^\circ$)

Part a: Using Law of Sines to find side a

Law of Sines: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

First, find angle $B$ or angle $A$. Wait, sum of angles: $A + B + C=180^\circ$, so $A + B=180 - 37.60 = 142.4^\circ$

From Law of Sines: $\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow\sin B=\frac{b\sin C}{c}$

Substitute $b = 50.900$, $c = 48.000$, $C = 37.60^\circ$

$\sin C=\sin(37.60^\circ)\approx0.612$

$\sin B=\frac{50.900\times0.612}{48.000}=\frac{31.1508}{48.000}\approx0.649$

So $B=\arcsin(0.649)\approx40.4^\circ$

Then angle $A=180-(37.60 + 40.4)=180 - 78=102^\circ$

Now, use Law of Sines to find $a$: $\frac{a}{\sin A}=\frac{c}{\sin C}$

$\sin A=\sin(102^\circ)\approx0.978$

$a=\frac{c\sin A}{\sin C}=\frac{48.000\times0.978}{0.612}=\frac{46.944}{0.612}\approx76.7$ mm? Wait, this is different.

Wait, maybe the angle of $37.60^\circ$ is angle $B$. Let's try:

Let $B = 37.60^\circ$, $b = 48.000$ mm, $c = 50.900$ mm

Then $\sin C=\frac{c\sin B}{b}=\frac{50.900\times\sin(37.60^\circ)}{48.000}$

$\sin(37.60^\circ)\approx0.612$

$\sin C=\frac{50.900\times0.612}{48.000}=\frac{31.1508}{48.000}\approx0.649$

$C=\arcsin(0.649)\approx40.4^\circ$

$A = 180-(37.60 + 40.4)=102^\circ$

Then $a=\frac{b\sin A}{\sin B}=\frac{48.000\times\sin(102^\circ)}{\sin(37.60^\circ)}=\frac{48.000\times0.978}{0.612}\approx76.7$ mm

Part b: Determine $\angle B$

If we assume $B = 37.60^\circ$ (from the problem statement's "37.60 degrees B and C" – maybe a misstatement, maybe one angle is $37.60^\circ$, let's say $B = 37.60^\circ$)

Part c: Determine $\angle C$

As calculated above, if $B = 37.60^\circ$, $A = 102^\circ$, then $C=180-(37.60 + 102)=40.4^\circ$

Wait, the problem statement's details are a bit unclear, but let's proceed with the Law of Sines.

Part a:

Let's assume that we have side $b = 50.900$ mm, side $c = 48.000$ mm, and angle $C = 37.60^\circ$

Using $\frac{b}{\sin B}=\frac{c}{\sin C}$

$\sin B=\frac{b\sin C}{c}=\frac{50.900\times\sin(37.60^\circ)}{48.000}$

$\sin(37.60^\circ)\approx0.612$

$\sin B=\frac{50.900\times0.612}{48.000}\approx\frac{31.1508}{48.000}\approx0.649$

$B\approx40.4^\circ$

Then angle $A = 180-(37.60 + 40.4)=102^\circ$

Now, $\frac{a}{\sin A}=\frac{c}{\sin C}$

$a=\frac{c\sin A}{\sin C}=\frac{48.000\times\sin(102^\circ)}{\sin(37.60^\circ)}$

$\sin(102^\circ)\approx0.978$, $\sin(37.60^\circ)\approx0.612$

$a=\frac{48.000\times0.978}{0.612}=\frac{46.944}{0.612}\approx76.7$ mm

Part b:

If we assume that the given angle is $B = 37.60^\circ$, then $\angle B = 37.60^\circ$

Part c:

$\angle C=180-(A + B)=180-(102 + 37.60)=40.4^\circ$ (approx)

Answer:

a. $\approx76.7$ mm (depending on the correct interpretation of the problem's angle - side correspondence)
b. $\approx37.60^\circ$ (if the given angle is $\angle B$)
c. $\approx40.4^\circ$ (approximate value)