Law of Sines and Cosines; Area of Triangles Mixed Practice ... - Free Printable

Problem Analysis and Solution

The worksheet involves solving triangles using the Law of Sines and Law of Cosines, as well as finding missing angles and areas. Let's solve each section step by step.

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Section A: Use an appropriate formula to find the missing lengths to 2 decimal places.

#### 1)
Given:
- \( \angle B = 83^\circ \)
- \( a = 4 \)
- \( c = 5 \)
- Find \( b \).

We use the Law of Cosines:
\[
b^2 = a^2 + c^2 - 2ac \cos B
\]
Substitute the given values:
\[
b^2 = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cdot \cos(83^\circ)
\]
\[
b^2 = 16 + 25 - 40 \cdot \cos(83^\circ)
\]
Using a calculator for \( \cos(83^\circ) \approx 0.1219 \):
\[
b^2 = 16 + 25 - 40 \cdot 0.1219
\]
\[
b^2 = 41 - 4.876
\]
\[
b^2 = 36.124
\]
\[
b = \sqrt{36.124} \approx 6.01
\]

Answer:
\[
b \approx 6.01
\]

#### 2)
Given:
- \( \angle A = 62^\circ \)
- \( \angle B = 41^\circ \)
- \( b = 4 \)
- Find \( a \).

First, find \( \angle C \):
\[
\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 62^\circ - 41^\circ = 77^\circ
\]

Use the Law of Sines:
\[
\frac{a}{\sin A} = \frac{b}{\sin B}
\]
\[
\frac{a}{\sin 62^\circ} = \frac{4}{\sin 41^\circ}
\]
\[
a = \frac{4 \cdot \sin 62^\circ}{\sin 41^\circ}
\]
Using a calculator for \( \sin 62^\circ \approx 0.8829 \) and \( \sin 41^\circ \approx 0.6561 \):
\[
a = \frac{4 \cdot 0.8829}{0.6561}
\]
\[
a \approx \frac{3.5316}{0.6561} \approx 5.38
\]

Answer:
\[
a \approx 5.38
\]

#### 3)
Given:
- \( \angle B = 120^\circ \)
- \( \angle C = 28^\circ \)
- \( b = 8.18 \)
- Find \( a \).

First, find \( \angle A \):
\[
\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 120^\circ - 28^\circ = 32^\circ
\]

Use the Law of Sines:
\[
\frac{a}{\sin A} = \frac{b}{\sin B}
\]
\[
\frac{a}{\sin 32^\circ} = \frac{8.18}{\sin 120^\circ}
\]
\[
a = \frac{8.18 \cdot \sin 32^\circ}{\sin 120^\circ}
\]
Using a calculator for \( \sin 32^\circ \approx 0.5299 \) and \( \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ \approx 0.8660 \):
\[
a = \frac{8.18 \cdot 0.5299}{0.8660}
\]
\[
a \approx \frac{4.336}{0.8660} \approx 5.01
\]

Answer:
\[
a \approx 5.01
\]

#### 4)
Given:
- \( \angle A = 90^\circ \)
- \( a = 7.51 \)
- \( b = 6.42 \)
- Find \( c \).

Since \( \angle A = 90^\circ \), this is a right triangle. Use the Pythagorean Theorem:
\[
c^2 = a^2 + b^2
\]
\[
c^2 = 7.51^2 + 6.42^2
\]
\[
c^2 = 56.4001 + 41.2164
\]
\[
c^2 = 97.6165
\]
\[
c = \sqrt{97.6165} \approx 9.88
\]

Answer:
\[
c \approx 9.88
\]

---

Section B: Use an appropriate formula to find the missing angles to 3 significant figures.

#### 1) Find angle PQR.
Given:
- \( PQ = 6 \)
- \( PR = 4.4 \)
- \( QR = 2.3 \)

Use the Law of Cosines to find \( \angle PQR \):
\[
\cos(\angle PQR) = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR}
\]
\[
\cos(\angle PQR) = \frac{6^2 + 2.3^2 - 4.4^2}{2 \cdot 6 \cdot 2.3}
\]
\[
\cos(\angle PQR) = \frac{36 + 5.29 - 19.36}{27.6}
\]
\[
\cos(\angle PQR) = \frac{21.93}{27.6}
\]
\[
\cos(\angle PQR) \approx 0.7953
\]
\[
\angle PQR = \cos^{-1}(0.7953) \approx 37.3^\circ
\]

Answer:
\[
\angle PQR \approx 37.3^\circ
\]

#### 2) The area of triangle DEF is 10 square units. Find angle FDE.
Given:
- \( DE = 3.85 \)
- \( EF = 7.63 \)
- Area = 10 square units

Use the area formula for a triangle:
\[
\text{Area} = \frac{1}{2}ab \sin C
\]
\[
10 = \frac{1}{2} \cdot 3.85 \cdot 7.63 \cdot \sin(\angle FDE)
\]
\[
10 = \frac{1}{2} \cdot 29.4155 \cdot \sin(\angle FDE)
\]
\[
10 = 14.70775 \cdot \sin(\angle FDE)
\]
\[
\sin(\angle FDE) = \frac{10}{14.70775} \approx 0.6801
\]
\[
\angle FDE = \sin^{-1}(0.6801) \approx 42.8^\circ
\]

Answer:
\[
\angle FDE \approx 42.8^\circ
\]

#### 3) Find angle LMN if angle LNM is obtuse.
Given:
- \( LN = 51 \)
- \( NM = 34 \)
- \( ML = 41 \)

First, use the Law of Cosines to find \( \angle LNM \):
\[
\cos(\angle LNM) = \frac{LN^2 + NM^2 - ML^2}{2 \cdot LN \cdot NM}
\]
\[
\cos(\angle LNM) = \frac{51^2 + 34^2 - 41^2}{2 \cdot 51 \cdot 34}
\]
\[
\cos(\angle LNM) = \frac{2601 + 1156 - 1681}{3468}
\]
\[
\cos(\angle LNM) = \frac{2076}{3468}
\]
\[
\cos(\angle LNM) \approx 0.600
\]
\[
\angle LNM = \cos^{-1}(0.600) \approx 53.1^\circ
\]

Since \( \angle LNM \) is obtuse, we take the supplementary angle:
\[
\angle LNM = 180^\circ - 53.1^\circ = 126.9^\circ
\]

Now, use the Law of Sines to find \( \angle LMN \):
\[
\frac{\sin(\angle LMN)}{LN} = \frac{\sin(\angle LNM)}{ML}
\]
\[
\frac{\sin(\angle LMN)}{51} = \frac{\sin(126.9^\circ)}{41}
\]
\[
\sin(\angle LMN) = \frac{51 \cdot \sin(126.9^\circ)}{41}
\]
Using a calculator for \( \sin(126.9^\circ) \approx 0.800 \):
\[
\sin(\angle LMN) = \frac{51 \cdot 0.800}{41}
\]
\[
\sin(\angle LMN) \approx \frac{40.8}{41} \approx 0.995
\]
\[
\angle LMN = \sin^{-1}(0.995) \approx 81.9^\circ
\]

Answer:
\[
\angle LMN \approx 81.9^\circ
\]

#### 4) YZT is a straight line. Find angle XZT.
Given:
- \( XY = 9.2 \)
- \( YZ = 6.4 \)
- \( XZ = 4.3 \)

Since \( YZT \) is a straight line, \( \angle XYZ = 180^\circ \). Use the Law of Cosines in \( \triangle XYZ \) to find \( \angle XYZ \):
\[
\cos(\angle XYZ) = \frac{XY^2 + YZ^2 - XZ^2}{2 \cdot XY \cdot YZ}
\]
\[
\cos(\angle XYZ) = \frac{9.2^2 + 6.4^2 - 4.3^2}{2 \cdot 9.2 \cdot 6.4}
\]
\[
\cos(\angle XYZ) = \frac{84.64 + 40.96 - 18.49}{117.76}
\]
\[
\cos(\angle XYZ) = \frac{107.11}{117.76}
\]
\[
\cos(\angle XYZ) \approx 0.909
\]
\[
\angle XYZ = \cos^{-1}(0.909) \approx 24.6^\circ
\]

Since \( YZT \) is a straight line:
\[
\angle XZT = 180^\circ - \angle XYZ = 180^\circ - 24.6^\circ = 155.4^\circ
\]

Answer:
\[
\angle XZT \approx 155.4^\circ
\]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Section A:} & 1) \, b \approx 6.01, \, 2) \, a \approx 5.38, \, 3) \, a \approx 5.01, \, 4) \, c \approx 9.88 \\
\text{Section B:} & 1) \, \angle PQR \approx 37.3^\circ, \, 2) \, \angle FDE \approx 42.8^\circ, \, 3) \, \angle LMN \approx 81.9^\circ, \, 4) \, \angle XZT \approx 155.4^\circ
\end{array}
}
\]