Question

What is the area of this triangle? (Image Attached)

Answer 1

$\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\ -----------------------------\\\\ \textit{so, let's find the missing side} \\\\\\ c=\sqrt{2.7^2+3.4^2-2(2.7)(3.4)cos(40^o)} \\\\\\ c=\sqrt{4.78542402433556327368}\implies c\approx 2.187561204706182220$

or we can round it, to say c = 2.19, so  hmm that's the missing side
now, we use Heron's Formula, which uses all 3 sides only

$\bf \textit{Heron's Area Formula}\\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} a=2.7\\ b=3.4\\ c\approx 2.19\\\\ s=\cfrac{a+b+c}{2} \end{cases}$

and that'd be the area of it

Answer 2

Answer:

The area of the triangle is 2.97 ft².

Step-by-step explanation:

We are given the triangle having sides 2.7 feet, 3.4 feet and the inclusive angle 40°.

We will find the length of the third side.

The law of cosines is given by $c^{2}=a^{2}+b^{2}-2ab\cos \theta$.

Substituting the values, we have,

$c^{2}=2.7^{2}+3.4^{2}-2\times 2.7\times 3.4\cos 40\\\\c^{2}=7.29+11.56-18.36\times 0.766\\\\c^{2}=18.85-14.06\\\\c^2=4.79\\\\c=2.2$

Thus, the length of the third side is 2.2 feet.

Now, the Heron's Area Formula is given by $A=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$.

So, $s=\frac{2.7+3.4+2.2}{2}\\\\s=\frac{8.3}{2}\\\\s=4.15$

Then, the area of the triangle is,

$A=\sqrt{4.15(4.15-2.7)(4.15-3.4)(4.15-2.2)}\\\\A=\sqrt{4.15\times 1.45\times 0.75\times 1.95}\\\\A=\sqrt{8.80059375}\\\\A=2.97$

Thus, the area of the triangle is 2.97 ft².