Question

1. determine the length of each indicated side to the nearest tenth of a centimetre.

a)
b)
c)
d)

Explanation:

Step1: Use cosine function for a)

In right - triangle $MPN$, $\cos\theta=\frac{adjacent}{hypotenuse}$. Given $\theta = 57^{\circ}$ and adjacent side $MP = 13.8$ cm, and we want to find hypotenuse $m$. So, $\cos(57^{\circ})=\frac{13.8}{m}$, then $m=\frac{13.8}{\cos(57^{\circ})}$.
$m=\frac{13.8}{0.5446}\approx25.3$ cm.

Step2: Use sine function for b)

In right - triangle $HKJ$, $\sin\theta=\frac{opposite}{hypotenuse}$. Given $\theta = 51^{\circ}$ and hypotenuse $j$ and opposite side $HJ = 6.2$ cm. So, $\sin(51^{\circ})=\frac{6.2}{j}$, then $j=\frac{6.2}{\sin(51^{\circ})}$.
$j=\frac{6.2}{0.7771}\approx8.0$ cm.

Step3: Use tangent function for c)

In right - triangle $EFG$, $\tan\theta=\frac{opposite}{adjacent}$. Given $\theta = 28^{\circ}$ and adjacent side $EF = 6.8$ cm and we want to find opposite side $f$. So, $\tan(28^{\circ})=\frac{f}{6.8}$, then $f = 6.8\times\tan(28^{\circ})$.
$f=6.8\times0.5317\approx3.6$ cm.

Step4: Use cosine function for d)

In right - triangle $BCD$, $\cos\theta=\frac{adjacent}{hypotenuse}$. Given $\theta = 42^{\circ}$ and hypotenuse $BD = 8.3$ cm and we want to find adjacent side $c$. So, $\cos(42^{\circ})=\frac{c}{8.3}$, then $c = 8.3\times\cos(42^{\circ})$.
$c=8.3\times0.7431\approx6.2$ cm.

Answer:

a) $m\approx25.3$ cm
b) $j\approx8.0$ cm
c) $f\approx3.6$ cm
d) $c\approx6.2$ cm