Problem 34

A triangle has a hypotenuse of length 3.00 m, one of its angles is 30.00.What are the lengths of:

(a) the side opposite the 30.00 angle?

SOLUTION:

The hypotenuse length is known so this problem is straightforward. Recall that sin of an angle is the opposite over the hypotenuse:

                                    sinq = opposite/hypotenuse

                                    opposite = (sinq) hypotenuse = (sin300)(3.00 m) = 1.50 m

(b) the side adjacent the 30.00 angle?

SOLUTION:

This is similar to the previous question but this time we use cosine.

                                    cosq = adjacent/hypotenuse

                                    adjacent = (cosq) hypotenuse = (cos 30.00) (3.00 m) = 2.60 m

Problem 36

In a certain right triangle, the two sides that are perpendicular to each other are 5.00 m and 7.00 m long. What is the length of the third side of the triangle?

SOLUTION:

The third side is the hypotenuse, which is not perpendicular to either of the other sides. We use the Pythagorean theorem to find the length of the hypotenuse.

Problem 42

Sphere 1 has an area A1 and a volume V1 and sphere 2 has an area A2 and a volume V2.If the radius of sphere 2 is double the radius of sphere 1, what is the ratio of :

(a) the areas A2/A1?

SOLUTION:

First, we need to know or look-up the surface area of a sphere: A = 4pR2

Therefore sphere 1: A1 = 4pR21 and sphere 2: A2 = 4pR22

But we know that the radius of sphere 2 is double that of sphere 1 or: R2 = 2R1

Therefore A2 = 4p (2R1)2 = 4p4R12 = 16pR21

So, the ratio is A2/A1 = (16pR21)/( 4pR21) = 4

 

(b) the volumes V2/V1?

SOLUTION:

We do it the same way we did the previous problem. V = (4/3)pR3

V1 = (4/3) pR13

V2 = (4/3) pR23 = (4/3) p (2R1)3 = (4/3) p (8R13)

V2/V1 = (4/3) p (8R13)/ (4/3) pR13 = 8