Question

name: hoshi 3.1 to 3.3 quiz
\\(\frac{a}{\sin a}=\frac{b}{\sin b}=\frac{c}{\sin c}\\) \\(c^{2}=a^{2}+b^{2}-2ab(\cos c)\\)

1. solve for the indicated angle (2):
2. solve for the indicated side (2):
3. sketch the following triangles, then solve for all side lengths and all interior angles. round to the nearest tenth (4 marks each).

a) in \\(\triangle abc,\angle c = 73^{\circ},\angle b = 32^{\circ},\\) and \\(a = 23\text{ cm}\\).
b) in \\(\triangle xyz,x = 12\text{ mm},z = 10.5\text{ mm}\\) and \\(\angle x = 60^{\circ}\\)

Explanation:

Step1: Find angle $\theta$ in the first triangle

Use the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. First find angle $A$ in the first - triangle.
We know that $\frac{37.1}{\sin\theta}=\frac{29.5}{\sin44^{\circ}}$. Then $\sin\theta=\frac{37.1\times\sin44^{\circ}}{29.5}$.
$\sin\theta=\frac{37.1\times0.6947}{29.5}=\frac{25.7734}{29.5}\approx0.8737$. So $\theta=\sin^{- 1}(0.8737)\approx61.0^{\circ}$.

Step2: Find side $b$ in the second triangle

First find angle $A$ in the second - triangle. Since the sum of angles in a triangle is $180^{\circ}$, $A = 180^{\circ}-72^{\circ}-43^{\circ}=65^{\circ}$.
Using the Law of Sines $\frac{b}{\sin B}=\frac{c}{\sin C}$, we have $\frac{b}{\sin72^{\circ}}=\frac{27.2}{\sin43^{\circ}}$. Then $b=\frac{27.2\times\sin72^{\circ}}{\sin43^{\circ}}$.
$\sin72^{\circ}\approx0.9511$, $\sin43^{\circ}\approx0.6820$. So $b=\frac{27.2\times0.9511}{0.6820}=\frac{25.8799}{0.6820}\approx37.9$ cm.

Step3a: Solve $\triangle ABC$

First find angle $A$: $A=180^{\circ}-73^{\circ}-32^{\circ}=75^{\circ}$.
Using the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.
$\frac{b}{\sin B}=\frac{a}{\sin A}$, so $b=\frac{a\sin B}{\sin A}=\frac{23\times\sin32^{\circ}}{\sin75^{\circ}}$. $\sin32^{\circ}\approx0.5299$, $\sin75^{\circ}\approx0.9659$. Then $b=\frac{23\times0.5299}{0.9659}=\frac{12.1877}{0.9659}\approx12.6$ cm.
$c=\frac{a\sin C}{\sin A}=\frac{23\times\sin73^{\circ}}{\sin75^{\circ}}$. $\sin73^{\circ}\approx0.9563$, $\sin75^{\circ}\approx0.9659$. Then $c=\frac{23\times0.9563}{0.9659}=\frac{21.9949}{0.9659}\approx22.8$ cm.

Step3b: Solve $\triangle XYZ$

Using the Law of Sines $\frac{x}{\sin X}=\frac{z}{\sin Z}$, so $\sin Z=\frac{z\sin X}{x}$.
$\sin Z=\frac{10.5\times\sin60^{\circ}}{12}$. $\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.8660$. Then $\sin Z=\frac{10.5\times0.8660}{12}=\frac{9.093}{12}\approx0.7578$. So $Z=\sin^{-1}(0.7578)\approx49.2^{\circ}$.
$Y = 180^{\circ}-60^{\circ}-49.2^{\circ}=70.8^{\circ}$.
$y=\frac{x\sin Y}{\sin X}=\frac{12\times\sin70.8^{\circ}}{\sin60^{\circ}}$. $\sin70.8^{\circ}\approx0.9440$, $\sin60^{\circ}\approx0.8660$. Then $y=\frac{12\times0.9440}{0.8660}=\frac{11.328}{0.8660}\approx13.1$ mm.

Answer:

1. $\theta\approx61.0^{\circ}$
2. $b\approx37.9$ cm

3a. $A = 75^{\circ}$, $b\approx12.6$ cm, $c\approx22.8$ cm
3b. $Y = 70.8^{\circ}$, $Z\approx49.2^{\circ}$, $y\approx13.1$ mm