10.3 Area of Triangle

Example:
Calculate the area of the triangle above.

Solution:
$\begin{array}{l}Area=\frac{1}{2}ab\sin C\\ Area=\frac{1}{2}(7)(4)\sin {40}^o\\ Area=9{\text{ cm}}^2\end{array}$

Example:

Diagram above shows a triangle ABC, where AC = 8 cm and ∠C = 32^o. Point D lies on straight line BC where BD = 10 cm and ∠ADB = 70^o . Calculate
(a) the length, in cm, of CD,
(b) the area, in cm² of ∆ ADC,
(c) the area, in cm² of ∆ ABC,
(d) the length, in cm, of AB.

Solution:
(a)
$\begin{array}{l}\angle ADC+{70}^o={180}^o\\ \angle ADC={110}^o\\ \angle CAD={180}^o-{110}^o-{32}^o\\ \angle CAD={38}^o\\ \\ \text{Using Sine Rule,}\\ \frac{8}{\sin {110}^o}=\frac{CD}{\sin {38}^o}\\ CD\sin {110}^o=8\sin {38}^o\\ CD\left(0.940\right)=8\left(0.616\right)\\ CD=5.243\end{array}$

(b)
$\begin{array}{l}\text{Area of }△ADC\\ =\frac{1}{2}\left(8\right)\left(5.243\right)\text{sin}{32}^o\\ =20.972\left(0.530\right)\\ =11.12{\text{ cm}}^2\end{array}$

(c)
$\begin{array}{l}\text{Area of }△ABC\\ =\frac{1}{2}\left(8\right)\left(10+5.243\right)\text{sin}{32}^o\\ =60.972\left(0.530\right)\\ =32.315{\text{ cm}}^2\end{array}$

(d)
$\begin{array}{l}\text{Use Cosine Rule for }△ABC,\\ A{B}^2={15.243}^2+8^2-2\left(15.243\right)\left(8\right)\cos {32}^o\\ A{B}^2=232.35+64-206.83\\ A{B}^2=89.52\\ AB=9.462\text{ cm}\end{array}$

Calculate the length of AC, x, in cm for the triangle above.

Solution:
$\begin{array}{l}{b}^2=a^2+c^2-2ac\cos B\\ x^2=4^2+7^2-2\left(4\right)\left(7\right)\cos {50}^{\circ }\\ x^2=16+49-56\left(0.6428\right)\\ x^2=65-35.997\\ x^2=29.003\\ x=5.385\text{ cm}\end{array}$

$\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos \angle BAC=\frac{7^2+6^2-8^2}{2(7)(6)}\\ \cos \angle BAC=0.25\\ \angle BAC={\cos }^{-1}0.25\\ \angle BAC={75.52}^o\end{array}$

10.1 The Sine Rule

$\begin{array}{l}\frac{AB}{\sin {60}^o}=\frac{4}{\sin {50}^o}\\ AB=\frac{4\times \sin {60}^o}{\sin {50}^o}\\ AB=4.522\text{ cm}\end{array}$

$\begin{array}{l}\frac{28}{\sin {54}^o}=\frac{26}{\sin \angle ACB}\\ \sin \angle ACB=\frac{26\times \sin {54}^o}{28}\\ \sin \angle ACB=0.7512\\ \angle ACB={48.7}^o\end{array}$