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Accepted Answer

### Part (a)
# Explanation:
## Step1: Identify the Law to Use
We have a triangle with sides $ AC = 9 $, $ AB = 6 $, $ BC = 14 $, and we need to find the angle $ x $ at $ A $. This is a case for the Law of Cosines, which is given by $ c^{2}=a^{2}+b^{2}-2ab\cos(C) $, where $ C $ is the angle opposite side $ c $. Here, side $ BC = 14 $ is opposite angle $ x $ at $ A $, so $ a = 9 $, $ b = 6 $, $ c = 14 $.
## Step2: Apply the Law of Cosines
Using the formula $ \cos(x)=\frac{AC^{2}+AB^{2}-BC^{2}}{2\cdot AC\cdot AB} $
Substitute the values: $ AC = 9 $, $ AB = 6 $, $ BC = 14 $
$ \cos(x)=\frac{9^{2}+6^{2}-14^{2}}{2\times9\times6}=\frac{81 + 36-196}{108}=\frac{117 - 196}{108}=\frac{- 79}{108}\approx - 0.7315 $
## Step3: Find the Angle
$ x=\cos^{-1}(-0.7315)\approx137.1^{\circ} $

### Part (b)
# Explanation:
## Step1: Find the Third Angle
In triangle $ ABC $, the sum of angles in a triangle is $ 180^{\circ} $. Given $ \angle A = 93^{\circ} $, $ \angle C=58^{\circ} $, so $ \angle B=180-(93 + 58)=29^{\circ} $
## Step2: Apply the Law of Sines
The Law of Sines is $ \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)} $. Here, $ AC = 16 $ (opposite $ \angle B = 29^{\circ} $), $ x $ is opposite $ \angle A=93^{\circ} $. So $ \frac{x}{\sin(93^{\circ})}=\frac{16}{\sin(29^{\circ})} $
## Step3: Solve for $ x $
$ x=\frac{16\times\sin(93^{\circ})}{\sin(29^{\circ})} $
$ \sin(93^{\circ})\approx0.9986 $, $ \sin(29^{\circ})\approx0.4848 $
$ x=\frac{16\times0.9986}{0.4848}\approx\frac{15.9776}{0.4848}\approx32.96\approx33 $

### Part (c)
# Explanation:
## Step1: Identify the Law to Use
We have a triangle with $ BC = 29 $, $ AC = 13 $, $ \angle C = 41^{\circ} $, and we need to find $ AB=x $. This is a case for the Law of Cosines. The Law of Cosines formula is $ AB^{2}=AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot\cos(C) $
## Step2: Apply the Law of Cosines
Substitute $ AC = 13 $, $ BC = 29 $, $ \angle C = 41^{\circ} $
$ x^{2}=13^{2}+29^{2}-2\times13\times29\times\cos(41^{\circ}) $
$ 13^{2}=169 $, $ 29^{2}=841 $, $ \cos(41^{\circ})\approx0.7547 $
$ x^{2}=169 + 841-2\times13\times29\times0.7547=1010-598.002=411.998 $
## Step3: Find $ x $
$ x=\sqrt{411.998}\approx20.3 $

### Part (d)
# Explanation:
## Step1: Apply the Law of Cosines
In triangle $ ABC $, $ AB = 10 $, $ AC = 17 $, $ BC = 21 $, and we need to find $ \angle C=x $. Using the Law of Cosines: $ \cos(x)=\frac{AC^{2}+BC^{2}-AB^{2}}{2\cdot AC\cdot BC} $
## Step2: Substitute the Values
$ AC = 17 $, $ BC = 21 $, $ AB = 10 $
$ \cos(x)=\frac{17^{2}+21^{2}-10^{2}}{2\times17\times21}=\frac{289+441 - 100}{714}=\frac{730 - 100}{714}=\frac{630}{714}\approx0.8823 $
## Step3: Find the Angle
$ x=\cos^{-1}(0.8823)\approx28.1^{\circ} $

### Part (e)
# Explanation:
## Step1: Apply the Law of Cosines
In triangle $ DWH $, $ DH $ is opposite $ W $, $ WH = 29 $, $ DW = 34 $, $ \angle H = 73^{\circ} $. Using the Law of Cosines: $ DW^{2}=DH^{2}+WH^{2}-2\cdot DH\cdot WH\cdot\cos(H) $. Let $ DH = x $, so $ 34^{2}=x^{2}+29^{2}-2\cdot x\cdot29\cdot\cos(73^{\circ}) $
## Step2: Rearrange the Equation
$ 1156=x^{2}+841-58x\times0.2924 $
$ x^{2}-17x + 841 - 1156 = 0 $
$ x^{2}-17x-315 = 0 $
## Step3: Solve the Quadratic Equation
Using the quadratic formula $ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} $, where $ a = 1 $, $ b=-17 $, $ c=-315 $
$ \Delta=b^{2}-4ac=(-17)^{2}-4\times1\times(-315)=289 + 1260 = 1549 $
$ x=\frac{17\pm\sqrt{1549}}{2}\approx\frac{17\pm39.36}{2} $
We take the positive root: $ x=\frac{17 + 39.36}{2}\approx28.18\approx28.2 $ (the negative root is discarded as length can't be negative)

### Part (f)
# Explanation:
## Step1: Apply the Law of Sines
In triangle $ DEF $, $ DE = 6.2 $, $ EF = 4.9 $, $ \angle E = 64^{\circ} $, $ x = DF $. By the Law of Sines: $ \frac{DF}{\sin(E)}=\frac{EF}{\sin(D)} $, but we can also use the Law of Cosines. $ DF^{2}=DE^{2}+EF^{2}-2\cdot DE\cdot EF\cdot\cos(E) $
## Step2: Substitute the Values
$ DE = 6.2 $, $ EF = 4.9 $, $ \cos(64^{\circ})\approx0.4384 $
$ x^{2}=6.2^{2}+4.9^{2}-2\times6.2\times4.9\times0.4384 $
$ 6.2^{2}=38.44 $, $ 4.9^{2}=24.01 $
$ x^{2}=38.44 + 24.01-2\times6.2\times4.9\times0.4384=62.45-26.73\approx35.72 $
## Step3: Find $ x $
$ x=\sqrt{35.72}\approx5.98\approx6.0 $

# Answers:
a) $ \approx137.1^{\circ} $

b) $ \approx33 $

c) $ \approx20.3 $

d) $ \approx28.1^{\circ} $

e) $ \approx28.2 $

f) $ \approx6.0 $

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Identify the Law to Use

Step2: Apply the Law of Cosines

Step3: Find the Angle

Part (b)

Step1: Find the Third Angle

Step2: Apply the Law of Sines

Step3: Solve for \( x \)

Part (c)

Step1: Identify the Law to Use

Step2: Apply the Law of Cosines

Step3: Find \( x \)

Part (d)

Answer: