Answer:
a) (i) 86.614 in
(ii) 47.6377 in
(iii) 1.584 m
b) width = 1.107 m (3 d.p.)
c) D = 1.135 m (3 d.p.)
Step-by-step explanation:
Part (a)
Question (i)
Given:
[tex]\implies \sf 2.2\: m = 2.2 \times 39.37 = 86.614\: in[/tex]
Question (ii)
Substitute the found value of D from part (i) into the given formula for D and solve for S:
[tex]\sf \implies D=\dfrac{S}{0.55}[/tex]
[tex]\sf \implies 86.614=\dfrac{S}{0.55}[/tex]
[tex]\implies \sf S = 86.614 \times 0.55[/tex]
[tex]\implies \sf S = 47.6377\: in[/tex]
Question (iii)
Given:
Substitute the given values into the given formula for M and solve for M:
[tex]\implies \sf M = E + D \times 0.22[/tex]
[tex]\implies \sf M = 1.1 + 2.2 \times 0.22[/tex]
[tex]\implies \sf M = 1.1 +0.484[/tex]
[tex]\implies \sf M = 1.584 \: m[/tex]
Part (b)
Given:
- S = 50 in
- Ratio of width to height = 16 : 9
Pythagoras Theorem
[tex]a^2+b^2=c^2[/tex]
where:
- a and b are the legs of the right triangle.
- c is the hypotenuse (longest side) of the right triangle.
Use Pythagoras Theorem to find the ratio of the diagonal S to the width and height:
[tex]\implies \sf 16^2+9^2=S^2[/tex]
[tex]\implies \sf 256+81=S^2[/tex]
[tex]\implies \sf S^2=337[/tex]
[tex]\implies \sf S=\sqrt{337}[/tex]
Therefore, the ratio of width to height to diagonal S of the TV is:
[tex]\sf w:h:S=16 : 9 : \sqrt{337}[/tex]
If S = 50 in then:
[tex]\implies \sf w:50=16:\sqrt{337}[/tex]
[tex]\implies \sf \dfrac{w}{50}=\dfrac{16}{\sqrt{337}}[/tex]
[tex]\implies \sf w=\dfrac{16}{\sqrt{337}} \times 50[/tex]
[tex]\implies \sf width=43.57877686..\:in[/tex]
To convert to meters, divide the width in inches by 39.37:
[tex]\implies \sf width=\dfrac{43.57877686}{39.37}=1.106903146\:m[/tex]
Therefore, the width of the TV is 1.107 m (3 d.p.).
Part (c)
If the TV is mounted in the position from part (aiii), where the midpoint of the TV is 1.584 m from the ground, and the height of the TV is 62.2cm, then:
[tex]\implies \textsf{Top of TV} \sf = 1.584 + \dfrac{62.2 \div 100}{2}[/tex]
[tex]\implies \textsf{Top of TV} \sf = 1.895\:m[/tex]
Therefore, the top of the TV from the floor is 1.895m.
As E = 1.1 m, the vertical distance between the eye level and the top of the TV is:
[tex]\implies \sf 1.895-1.1=0.795\:m[/tex]
To find D, model as a right triangle (see attached) and use the tan trigonometric ratio to find the shortest value of D.
Tan trigonometric ratio
[tex]\sf \tan(\theta)=\dfrac{O}{A}[/tex]
where:
- [tex]\theta[/tex] is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
Given:
- [tex]\theta[/tex] = 35°
- O = 0.795 m
- A = D m
Substitute the values into the formula and solve for D:
[tex]\implies \sf \tan(35^{\circ})=\dfrac{0.795}{D}[/tex]
[tex]\implies \sf D=\dfrac{0.795}{\tan(35^{\circ})}[/tex]
[tex]\implies \sf D=1.135377665\:m[/tex]
Therefore, the shortest distance that the sofa should be placed from the wall is 1.135 m (3 d.p.).
