A triangle has the sides of lengths 5 and 8. The measure of the angle included between these two sides is 105 degrees. calculate the length of the third side and the area of the triangle, to the nearest tenth.
how do I draw this triangle?which side is a which side is b here? How?
+245 It doesn't matter which one is a or b. A and B are just generics labels. Make whatever side you want a and the other b.
Just make sure you have an obtuse looking angle. Like this one. And label it 105 degrees.
Next, make the side with the bigger value the longer of the two sides. Whether that is a or b it doesn't matter.
Also, make sure that c is the longest as the pic above shows.
Here is what I drew.
+245 Sources: http://www.mathwarehouse.com/geometry/triangles/area/herons-formula-triangle-area.php
https://www.mathsisfun.com/algebra/trig-cosine-law.html
Using the law of Cosines:
\(C^2 = A^2 +B^2 - 2ABcos(c)\)
Where C is the unknown side
A is one the known sides
B is the other known side
c is the angle between the two known sides
Now plug in values to find C
\(C^2 = 5^2 + 8^2 - 2(5)(8)cos105\) = \(109.7055 \)
\(\sqrt(C^2)= C\)
\(\sqrt(109.7055)\) = \(10.474 = 10.5 = C \)
Finding the area (using Heron's Formula);
\(Area= \sqrt(S(S-A)(S-B)(S-C)\)
where S is the semi-perimeter (half the perimeter)
A is one of the sides
B is another side
C is the final side
Now to plug it in:
The area is 19.9
