QUESTION 1
We can use the cosine rule to find the missing side length.
Recall that the cosine rule for a triangle with sides a,b,c and an included angle A is
[tex]a^2=b^2+c^2-2bc\cos A[/tex]
Let the missing side length in the triangle with sides 6, 9 and the included angle of [tex]37\degree[/tex] be [tex]a[/tex] units.
We then substitute the values into the cosine rule to obtain;
[tex]a^2=6^2+9^2-2(6)(9)\cos 37\degree[/tex]
[tex]a^2=36+81-108\cos 37\degree[/tex]
[tex]a^2=30.747[/tex]
[tex]\Rightarrow a=\sqrt{30.747}[/tex]
[tex]\Rightarrow a=\sqrt{30.747}[/tex]
[tex]\Rightarrow a=5.5[/tex] units to the nearest tenth.
QUESTION 2
We again use the cosine rule: [tex]a^2=b^2+c^2-2bc\cos A[/tex]
We substitute the given values to obtain;
[tex]a^2=11^2+13^2-2(11)(13)\cos 108\degree[/tex]
[tex]a^2=121+169-286\cos 108\degree[/tex]
[tex]a^2=378.379[/tex]
[tex]\Rightarrow a=\sqrt{378.379}[/tex]
[tex]\Rightarrow a=19.5[/tex] to the nearest tenth
QUESTION 3
We again use the cosine rule :
[tex]|BA|^2=(\frac{1}{2})^2+(\frac{1}{3})^2-2(\frac{1}{2})(\frac{1}{3})\cos 100\degree[/tex]
[tex]|BA|^2=0.418899[/tex]
[tex]|BA|=\sqrt{0.418899}[/tex]
[tex]|BA|=0.65[/tex] to the nearest hundredth