Answer:
7) A = 137.0°, B = 26.0°, C = 17.0°
8) B = 145°, a = 15.0 in, b = 22.9 in
Step-by-step explanation:
Question 7
When we solve a triangle, we need to find all of its unknown angles and side lengths.
In the case of triangle ABC, all three side lengths are given, so, to solve triangle ABC, we need to find the measure of angles A, B and C.
To find the measure of one of the angles of triangle ABC, we can use the Law of Cosines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Cosine Rule (for finding angles)}} \\\\\cos(C)=\dfrac{a^2+b^2-c^2}{2ab}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$C$ is the angle.}\\ \phantom{ww}\bullet\;\textsf{$a$ and $b$ are the sides adjacent the angle.}\\ \phantom{ww}\bullet\;\textsf{$c$ is the side opposite the angle.}\end{array}}[/tex]
In this case:
- a = 14 cm
- b = 9 cm
- c = 6 cm
Substitute the side lengths into the equation and solve for angle C:
[tex]\cos(C)=\dfrac{14^2+9^2-6^2}{2(14)(9)}\\\\\\\cos(C)=\dfrac{196+81-36}{2(14)(9)}\\\\\\\cos(C)=\dfrac{241}{252}\\\\\\C=\cos^{-1}\left(\dfrac{241}{252}\right)\\\\\\C=16.991286936969...^{\circ}\\\\\\C\approx 17.0^{\circ}[/tex]
Now, we can use the Law of Sines to find another angle.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
In this case:
[tex]\dfrac{\sin A}{14}=\dfrac{\sin B}{9}=\dfrac{\sin \left(\cos^{-1}\left(\dfrac{241}{252}\right)\right)}{6}[/tex]
Solve for angle B:
[tex]\dfrac{\sin B}{9}=\dfrac{\sin \left(\cos^{-1}\left(\dfrac{241}{252}\right)\right)}{6}\\\\\\\sin B=\dfrac{9\sin \left(\cos^{-1}\left(\dfrac{241}{252}\right)\right)}{6}\\\\\\\sin B=0.438339411768...\\\\\\B=\sin^{-1}(0.438339411768...)\\\\\\B=25.99797699249...^{\circ}\\\\\\B\approx 26.0^{\circ}[/tex]
Since the interior angles of a triangle add up to 180°, to determine the measure of angle A, subtract the measures of angles B and C from 180°:
[tex]A = 180^{\circ}-B-C\\\\A = 180^{\circ}-25.99797699249...^{\circ}-16.991286936969...^{\circ}\\\\A = 137.01073607...^{\circ}\\\\A \approx 137.0^{\circ}[/tex]
Therefore, the angles of triangle ABC rounded to the nearest tenth are:
[tex]\Large\boxed{\boxed{\begin{array}{l}A = 137.0^{\circ}\\B = 26.0^{\circ}\\C = 17.0^{\circ}\end{array}}}[/tex]
[tex]\dotfill[/tex]
Question 8
For triangle ABC, we are given that angle C measures 13°, angle A measures 22° and side c measures 9 inches.
To solve the triangle, we need to find the measure of angle B and the lengths of sides a and b.
Since the interior angles of a triangle add up to 180°, to determine the measure of angle B, subtract the measures of angles A and C from 180°:
[tex]B = 180^{\circ}-A-C\\\\B=180-13^{\circ}-22^{\circ}\\\\B=145^{\circ}[/tex]
To find the unknown side lengths a and b, we can use the Law of Sines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
In this case:
[tex]\dfrac{a}{\sin 22^{\circ}}=\dfrac{b}{\sin 145^{\circ}}=\dfrac{9}{\sin 13^{\circ}}[/tex]
Solve for side a:
[tex]\dfrac{a}{\sin 22^{\circ}}=\dfrac{9}{\sin 13^{\circ}}\\\\\\a=\dfrac{9\sin 22^{\circ}}{\sin 13^{\circ}}\\\\\\a=14.987524066411\\\\\\ a\approx 15.0\; \sf in[/tex]
Solve for side b:
[tex]\dfrac{b}{\sin 145^{\circ}}=\dfrac{9}{\sin 13^{\circ}}\\\\\\b=\dfrac{9\sin 145^{\circ}}{\sin 13^{\circ}}\\\\\\b=22.948049486660...\\\\\\b\approx 22.9\; \sf in[/tex]
Therefore, the unknown angle and side lengths of triangle ABC rounded to the nearest tenth are:
[tex]\Large\boxed{\boxed{\begin{array}{l}B = 145^{\circ}\\a=15.0\; \textsf{in}\\b = 22.9\; \textsf{in}\end{array}}}[/tex]
