Hello aju051000,
1.) I would first draw a diagram:
View attachment 1246
We are given the following:
$$a=4\text{ ft},\,b=8\text{ ft},\,\theta=60^{\circ}$$
So, we need to find the $s$ and $w$. We may use:
$$\tan\left(60^{\circ} \right)=\frac{a}{w/2}=\frac{2a}{w}$$
$$w=2a\cot\left(60^{\circ} \right)=\frac{8}{\sqrt{3}}\text{ ft}$$
We should recognize that the triangular ends of the tent are equilateral (and so $s=w$), but if we didn't we could write:
$$\sin\left(60^{\circ} \right)=\frac{a}{s}$$
$$s=a\csc\left(60^{\circ} \right)=\frac{8}{\sqrt{3}}\text{ ft}$$
So, to find the surface area of the tent, we see that there are two congruent triangles and 3 congruent rectangles.
The area of the two triangles is:
$$A_T=2\cdot\frac{1}{2}\cdot\frac{8}{\sqrt{3}}\text{ ft}\cdot4\text{ ft}=\frac{32}{\sqrt{3}}\text{ ft}^2$$
The area of the three rectangles is:
$$A_R=3\cdot\frac{8}{\sqrt{3}}\text{ ft}\cdot8\text{ ft}=64\sqrt{3}\text{ ft}^2$$
Hence, the total surface area $A$ of the tent is:
$$A=A_T+A_R=\frac{32}{\sqrt{3}}\text{ ft}^2+64\sqrt{3}\text{ ft}^2=\frac{224}{\sqrt{3}}\text{ ft}^2$$
2.) Again, let's first draw a diagram:
View attachment 1247
We can now easily see that $\alpha$ and $\beta$ are complementary, hence:
$$\alpha+\beta=90^{\circ}$$
$$\alpha=90^{\circ}-\beta$$
With $\beta=62^{\circ}$, we find:
$$\alpha=(90-62)^{\circ}=28^{\circ}$$
