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Question:
Give step-by-step solution with explanation and final answer:
rn
i le ABC. (Round the length to three BE
ER pr ——
ER =
E i —
| g ==
= 10 23 =
= | Need Help? [HEE] msl Ee
FE FE
ee een ————
EL Be ca i
Ee. He si
fe Bn
the Indicated id x. (Use ithe th Law of Sines or th Low of Coss, 1 eproprit. Asse. 10 nd
Ee fi
1 c LL
3 Hid
he LL
Nad Help? (Em) (an
3 = Ii!
o SEER TO
eT —————————————] i
| aes, sous cus 4
3
rm [== a ie
i
i
pep ——T i
= a pr
Ee 14
N |
!
$5 iH |
4] i i
HH BERL
Pill Bi
x H
BH ELI
|
i !
ih |
i
A :
i |
i
| A.
L i
H 3
ER
- i
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Asked by: gvr
Created at: 2025-07-29 11:12:48
Asked by: gvr
Created at: 2025-07-29 11:12:48
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Let's solve each problem step by step!
---
### **1. Triangle ABC (First Image, First Problem)**
Given:
- \( AC = 10 \)
- \( BC = 23 \)
- \( \angle C = 120^\circ \)
Let \( AB = c \), \( \angle A = LA \), \( \angle B = LB \).
#### **Step 1: Find side \( c \) using Law of Cosines**
\[
c^2 = 10^2 + 23^2 - 2 \cdot 10 \cdot 23 \cdot \cos(120^\circ)
\]
\[
\cos(120^\circ) = -0.5
\]
\[
c^2 = 100 + 529 - 2 \cdot 10 \cdot 23 \cdot (-0.5)
\]
\[
c^2 = 629 + 230 = 859
\]
\[
c = \sqrt{859} \approx 29.323
\]
---
#### **Step 2: Find Angle \( LA = \angle A \) using Law of Sines**
\[
\frac{\sin(A)}{23} = \frac{\sin(120^\circ)}{29.323}
\]
\[
\sin(120^\circ) = 0.8660
\]
\[
\sin(A) = \frac{23 \times 0.8660}{29.323} = \frac{19.918}{29.323} \approx 0.6795
\]
\[
A = \arcsin(0.6795) \approx 42.8^\circ
\]
---
#### **Step 3: Find Angle \( LB = \angle B \)**
\[
\angle B = 180^\circ - 120^\circ - 42.8^\circ = 17.2^\circ
\]
---
#### **Final Answers:**
- \( c = 29.323 \)
- \( LA = 42.8^\circ \)
- \( LB = 17.2^\circ \)
---
### **2. Triangle ABC (Second Image, Second Problem)**
Given:
- \( a = 16 \), \( b = 17 \), \( c = 28 \)
Let \( LA = \angle A \), \( LB = \angle B \), \( LC = \angle C \).
#### **Step 1: Use Law of Cosines to find \( \angle A \):**
\[
a^2 = b^2 + c^2 - 2bc\cos(A)
\]
\[
16^2 = 17^2 + 28^2 - 2 \times 17 \times 28 \cos(A)
\]
\[
256 = 289 + 784 - 2 \times 17 \times 28 \cos(A)
\]
\[
256 = 1073 - 952 \cos(A)
\]
\[
952\cos(A) = 1073 - 256 = 817
\]
\[
\cos(A) = \frac{817}{952} \approx 0.8582
\]
\[
A = \arccos(0.8582) \approx 30.5^\circ
\]
---
#### **Step 2: Use Law of Cosines to find \( \angle B \):**
\[
b^2 = a^2 + c^2 - 2ac\cos(B)
\]
\[
17^2 = 16^2 + 28^2 - 2 \times 16 \times 28 \cos(B)
\]
\[
289 = 256 + 784 - 896 \cos(B)
\]
\[
289 = 1040 - 896 \cos(B)
\]
\[
896\cos(B) = 1040 - 289 = 751
\]
\[
\cos(B) = \frac{751}{896} \approx 0.8382
\]
\[
B = \arccos(0.8382) \approx 33.3^\circ
\]
---
#### **Step 3: Find \( \angle C \):**
\[
C = 180^\circ - 30.5^\circ - 33.3^\circ = 116.2^\circ
\]
---
#### **Final Answers:**
- \( LA = 30.5^\circ \)
- \( LB = 33.3^\circ \)
- \( LC = 116.2^\circ \)
---
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