Chapter 12.9, Problem 19WE

(a)

To determine

Justify $a^2+b^2-c^2=2ab\cos C$ .

Explanation of Solution

Calculation:

Consider the triangle:

  

Law of Cosine:

  $\begin{array}{l}{c}^2=a^2+b^2-2ab\cos C\\ \Rightarrow a^2+b^2-c^2-2ab\cos C=\mathrm{0........}[{\text{subtract c}}^2\text{ from each side]}\\ \Rightarrow a^2+b^2-c^2=2ab\cos C\mathrm{..............}[\text{add 2abcosC each side]}\end{array}$

Hence proved.

(b)

To determine

Justify ${(a^2+b^2-c^2)}^2=4a^2{b}^2{\cos }^{2}C=4a^2{b}^2(1-{\sin }^{2}C)$ .

Explanation of Solution

Calculation:

From part (a)

  $\begin{array}{l}{a}^2+b^2-c^2=2ab\cos C\\ \Rightarrow {\left(a^2+b^2-c^2\right)}^2=4a^2{b}^2{\cos }^{2}C\mathrm{...........}[\text{square each side]}\\ \Rightarrow {\left(a^2+b^2-c^2\right)}^2=4a^2{b}^2(1-{\sin }^{2}C)\mathrm{....}[{\sin }^2\theta +{\cos }^2\theta =1]\end{array}$

Hence proved.

(c)

To determine

Justify ${(a^2+b^2-c^2)}^2=4a^2{b}^2-4a^2{b}^2{\sin }^{2}C=4a^2{b}^2-16K^2$ .

Explanation of Solution

Calculation:

From part (b)

  ${\left(a^2+b^2-c^2\right)}^2=4a^2{b}^2(1-{\sin }^{2}C)$

From the result proved in Q17 ,

  $\begin{array}{l}{K}^2=\frac{1}{4}{a}^2{b}^2{\sin }^{2}C\\ \Rightarrow 16K^2=16\left(\frac{1}{4}{a}^2{b}^2{\sin }^{2}C\right)\mathrm{............}[\text{multiply 16 on each side]}\\ \Rightarrow 16K^2=4a^2{b}^2{\sin }^{2}C\end{array}$

So,

  $\begin{array}{l}{\left(a^2+b^2-c^2\right)}^2=4a^2{b}^2(1-{\sin }^{2}C)\\ \Rightarrow {\left(a^2+b^2-c^2\right)}^2=4a^2{b}^2-4a^2{b}^2{\sin }^{2}C\\ \Rightarrow {\left(a^2+b^2-c^2\right)}^2=4a^2{b}^2-16K^2\end{array}$

Hence proved.

(d)

To determine

Justify $16K^2=4a^2{b}^2-{(a^2+b^2-c^2)}^2$ .

Explanation of Solution

Calculation:

From part (c)

  $\begin{array}{}$ {(a^2+b^2-c^2)}^2=4a^2{b}^2-16K^2$$ \Rightarrow {(a^2+b^2-c^2)}^2+16K^2=4a^2{b}^2\mathrm{.............}[\text{add }16K^2\text{ each side]}$$ \Rightarrow 16K^2=4a^2{b}^2-{(a^2+b^2-c^2)}^2\mathrm{.............}[\text{subtract }{(a^2+b^2-c^2)}^2\text{ from each side]}$\end{array}$

Hence proved.

(e)

To determine

Justify $16K^2=\left[2ab+(a^2+b^2-c^2)\right]\left[2ab-(a^2+b^2-c^2)\right]$ .

Explanation of Solution

Calculation:

From part (d)

  $\begin{array}{}$ 16K^2=4a^2{b}^2-{(a^2+b^2-c^2)}^2$$ \Rightarrow 16K^2={(2ab)}^2-{(a^2+b^2-c^2)}^2$$ \Rightarrow 16K^2=\left[2ab+(a^2+b^2-c^2)\right]\left[2ab-(a^2+b^2-c^2)\right]\mathrm{.....}[a^2-b^2=(a+b)(a-b)]$\end{array}$

Hence proved.

(f)

To determine

Justify $16K^2=\left[{\left(a+b\right)}^2-c^2\right]\left[c^2-{(a-b)}^2\right]$ .

Explanation of Solution

Calculation:

From part (e)

  $\begin{array}{l}16K^2=\left[2ab+(a^2+b^2-c^2)\right]\left[2ab-(a^2+b^2-c^2)\right]\\ \Rightarrow 16K^2=\left[(a^2+2ab+b^2)-c^2)\right]\left[c^2-(a^2-2ab+b^2)\right]\mathrm{.....}[\text{associative property]}\\ $ \Rightarrow 16K^2=\left[{\left(a+b\right)}^2-c^2\right]\left[c^2-{(a-b)}^2\right]\mathrm{.............................}[{(a\pm b)}^2=a^2\pm 2ab+b^2]$\end{array}$

Hence proved.

(g)

To determine

Justify $16K^2=\left[\left(a+b\right)+c\right]\left[\left(a+b\right)-c\right]\left[c+(a-b)\right]\left[c-(a-b)\right]$ .

Explanation of Solution

Calculation:

From part (f)

  $\begin{array}{l}16K^2=\left[{\left(a+b\right)}^2-c^2\right]\left[c^2-{(a-b)}^2\right]\\ \Rightarrow 16K^2=\left[\left(a+b\right)+c\right]\left[\left(a+b\right)-c\right]\left[c+(a-b)\right]\left[c-(a-b)\right]\mathrm{....}[a^2-b^2=(a+b)(a-b)]\end{array}$

Hence proved.

(h)

To determine

Justify $16K^2=\left[2s\right]\left[2(s-c)\right]\left[2(s-b)\right]\left[2(s-a)\right]$ .

Explanation of Solution

Calculation:

From part (g)

  $16K^2=\left[\left(a+b\right)+c\right]\left[\left(a+b\right)-c\right]\left[c+(a-b)\right]\left[c-(a-b)\right]$

From Question 18 ,

  $2s=a+b+c\text{ , }2(s-a)=-a+b+c\text{ , }2(s-b)=a-b+c\text{ and }2(s-c)=a+b-c$

So,

  $\begin{array}{l}16K^2=\left[\left(a+b\right)+c\right]\left[\left(a+b\right)-c\right]\left[c+(a-b)\right]\left[c-(a-b)\right]\\ \Rightarrow 16K^2=\left[a+b+c\right]\left[a+b-c\right]\left[a-b+c\right]\left[-a+b+c\right]\mathrm{......}[\text{associative property]}\\ \Rightarrow 16K^2=\left[2s\right]\left[2(s-c)\right]\left[2(s-b)\right]\left[2(s-a)\right]\end{array}$

Hence proved.

(i)

To determine

Justify $K=\sqrt{s(s-a)(s-b)(s-c)}$ .

Explanation of Solution

Calculation:

From part (h)

  $\begin{array}{l}16K^2=\left[2s\right]\left[2(s-c)\right]\left[2(s-b)\right]\left[2(s-a)\right]\\ \Rightarrow 16K^2=16\left[s\right]\left[(s-c)\right]\left[(s-b)\right]\left[(s-a)\right]\\ \Rightarrow K^2=s(s-a)(s-b)(s-c)\mathrm{........................}[\text{divide each side by 16 and use associative property]}\\ \Rightarrow K=\sqrt{s(s-a)(s-b)(s-c)}\mathrm{.......................}[\text{take square root each side]}\end{array}$

Hence proved.