As given by the question
There are given that the vertex angle is 42 degrees and the base is 6 inches.
Now,
If the vertex angle of the isosceles triangle is 42 degrees, so each of the base angles is:
[tex]\begin{gathered} \frac{180-42}{2}=\frac{138}{2} \\ =69 \end{gathered}[/tex]And,
The base of the triangle is 6 inches
So, the altitude of the triangle is:
[tex]\begin{gathered} (\frac{6}{2})\times\tan 69^{\circ}=3\times\tan 69^{\circ} \\ =7.81\text{ inches} \end{gathered}[/tex]Then,
The area of the triangle is:
[tex]\begin{gathered} \text{Area of triangle=}\frac{6\times7.81}{2} \\ \text{Area of triangle}=23.43 \end{gathered}[/tex]Hence, the area of the triangle is 23.43.
