Answer:
14. 46.2 ft (nearest tenth)
15. 95.5° (nearest tenth)
Step-by-step explanation:
Question 14
The diagonal of the square (from home plate to second base) is:
[tex]\sqrt{65^2+65^2} =91.92388155...[/tex]
Half of this is 45.9619... ft.
Therefore, ∠ABC is not 90° and so triangle ABC is NOT a right triangle.
To determine length AB we must use the cosine rule:
[tex]c^2=a^2+b^2-2ab \cos C[/tex]
(where a and b are the sides, C is the include angle, and c is the side opposite the angle)
Given:
- a = BC = 41.5 ft
- b = AC = 65 ft
- c = AB
- C = 45°
Substituting given values into the formula:
[tex]\implies c^2=41.5^2+65^2-2(41.5)(65) \cos (45)[/tex]
[tex]\implies c^2=1722.25+4225-5395 \cdot \dfrac{\sqrt{2} }{2}[/tex]
[tex]\implies c^2=2132.408915...[/tex]
[tex]\implies c=\pm\sqrt{2132.408915...}[/tex]
[tex]\implies c=\pm46.17801333...[/tex]
As distance is positive, c = 46.2 ft (nearest tenth)
Question 15
We need to find angle B.
Use the sine rule:
[tex]\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}[/tex]
(where A, B and C are the angles, and a, b and c are the sides opposite the angles)
IMPORTANT: As angle B is obtuse (more than 90° and less than 180°), the sine of an obtuse angle = sine of its supplement
Therefore, sin B = sin (180 - B)
Given:
- C = 45°
- c = 46.17801333...
- b = AC = 65
- sin B = sin (180 - B)
[tex]\implies \dfrac{\sin(180-B)}{65}=\dfrac{\sin(45)}{46.17801333...}[/tex]
[tex]\implies \sin(180-B)=\dfrac{65 \sin(45)}{46.17801333...}[/tex]
[tex]\implies 180-B=84.45515638...[/tex]
[tex]\implies B=180-84.45515638...[/tex]
[tex]\implies B=95.54484362...[/tex]
Therefore, B = 95.5° (nearest tenth)
** Note: if you use the rounded solution for c, where c = 46.2, in this calculation, then angle B will be 95.8° to the nearest tenth**
