Answer:
Find below the calculations of the two areas, each with two methods. The results are:
[tex]Area=5000\sqrt{3}units^2[/tex]
[tex]Area=14,530m^2[/tex]
Explanation:
A) Method 1
When you are not given the height, but you are given two sides and the included angle between the two sides, you can use this formula:
[tex]Area=side_1\times side_2\times sin(\alpha)[/tex]
Where, [tex]\alpha[/tex] is the measure of the included angle.
1. Upper triangle:
[tex]side_1=200units\\ \\ side_2=100units\\ \\ \alpha =60\º\\ \\ Area=200units\times 100units\times sin(60\º)/2\\ \\ Area=5000\sqrt{3}units^2[/tex]
2. Lower triangle:
[tex]side_1=231m\\ \\ side_2=150m\\ \\ \alpha =123\º\\ \\ Area=231m\times 150m\times sin(123\º)/2\\ \\ Area=14,529.96m^2\approx14,530m^2[/tex]
B) Method 2
You can find the height of the triangle using trigonometric properties, and then use the very well known formula:
[tex]Area=(1/2)\times base\times height[/tex]
Use it for both triangles.
3. Upper triangle:
The trigonometric ratio that you can use is:
[tex]sine(\alpha)=opposite\text{ }leg/hypotenuse[/tex]
Notice the height is the opposite leg to the angle of 60º, and the side that measures 100 units is the hypotenuse of that right triangle. Then:
[tex]sin(60\º)=height/100units\\ \\ height=sin(60\º)\times100units\\ \\ height=50\sqrt{3}units[/tex]
[tex]Area=(1/2)\times base\times height=(1/2)\times 200units\times 50\sqrt{3}units=5,000\sqrt{3}units^2[/tex]
3. Lower triangle:
[tex]sin(180\º-123\º)=height/231m\\ \\ height=sin(57\º)\times 231m\\ \\ height=193.7329m^2[/tex]
[tex]Area=(1/2)\times base\times height=(1/2)\times 150m\times 193.7329m^2\\\\ Area=14,529.96m^2\approx 14,530m^2[/tex]
