Question

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

a) in △abc, ∠c = 66°, a = 32 cm, and b = 23 cm.
b) in △xyz, x = 12 mm, y = 10.5 mm and z = 14 mm

Explanation:

Step 1: For question 4 (Using the Law of Cosines)

The Law of Cosines formula for finding an angle in a triangle with sides \(a\), \(b\), \(c\) is \(\cos P=\frac{q^{2}+r^{2}-p^{2}}{2qr}\). Here, \(p = 6.2\), \(q=2.3\), \(r = 5.9\).
\[

$$\begin{align*} \cos P&=\frac{2.3^{2}+5.9^{2}-6.2^{2}}{2\times2.3\times5.9}\\ &=\frac{5.29 + 34.81- 38.44}{27.14}\\ &=\frac{1.66}{27.14}\\ &\approx0.0612 \end{align*}$$

\]

Step 2: Find angle \(P\)

\(P=\cos^{-1}(0.0612)\approx 86.5^{\circ}\)

Step 3: For question 5 (Using the Law of Cosines)

The Law of Cosines formula for side \(x\) in \(\triangle XYZ\) is \(x^{2}=y^{2}+z^{2}-2yz\cos X\). Here, \(y = 15\), \(z = 18\), \(X=46^{\circ}\), \(\cos X=\cos46^{\circ}\approx0.6947\)
\[

$$\begin{align*} x^{2}&=15^{2}+18^{2}-2\times15\times18\times0.6947\\ &=225+324 - 324\times0.6947\\ &=549- 225.0828\\ &=323.9172 \end{align*}$$

\]

Step 4: Find side \(x\)

\(x=\sqrt{323.9172}\approx17.99\approx18\mathrm{cm}\)

Step 5: For question 7a (Using the Law of Cosines and Law of Sines)

First, find side \(c\) using the Law of Cosines \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Given \(a = 32\), \(b = 23\), \(C = 66^{\circ}\), \(\cos C=\cos66^{\circ}\approx0.4067\)
\[

$$\begin{align*} c^{2}&=32^{2}+23^{2}-2\times32\times23\times0.4067\\ &=1024+529-597.9328\\ &=955.0672 \end{align*}$$

\]
\(c=\sqrt{955.0672}\approx31\mathrm{cm}\)
Then, use the Law of Sines \(\frac{\sin A}{a}=\frac{\sin C}{c}\) to find \(\sin A\), \(\sin A=\frac{a\sin C}{c}=\frac{32\times\sin66^{\circ}}{31}\approx\frac{32\times0.9135}{31}\approx0.947\), \(A=\sin^{-1}(0.947)\approx71^{\circ}\)
\(B=180^{\circ}-A - C=180^{\circ}-71^{\circ}-66^{\circ}=43^{\circ}\)

Step 6: For question 7b (Using the Law of Cosines and Law of Sines)

First, find \(\cos X=\frac{y^{2}+z^{2}-x^{2}}{2yz}=\frac{10.5^{2}+14^{2}-12^{2}}{2\times10.5\times14}=\frac{110.25 + 196-144}{294}=\frac{162.25}{294}\approx0.552\)
\(X=\cos^{-1}(0.552)\approx56.5^{\circ}\)
Using the Law of Sines \(\frac{\sin Y}{y}=\frac{\sin X}{x}\), \(\sin Y=\frac{y\sin X}{x}=\frac{10.5\times\sin56.5^{\circ}}{12}\approx\frac{10.5\times0.8346}{12}\approx0.73\)
\(Y=\sin^{-1}(0.73)\approx47^{\circ}\)
\(Z=180^{\circ}-X - Y=180^{\circ}-56.5^{\circ}-47^{\circ}=76.5^{\circ}\)

Answer:

1. Angle \(P\approx86.5^{\circ}\)
2. Side \(x\approx18\mathrm{cm}\)

7a. \(c\approx31\mathrm{cm}\), \(A\approx71^{\circ}\), \(B = 43^{\circ}\)
7b. \(X\approx56.5^{\circ}\), \(Y\approx47^{\circ}\), \(Z\approx76.5^{\circ}\)