Question

Only need help with parts (b) and (c)!! thank you!!

Answer 1

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:

• 1 m = 39.37 in

$\implies \sf 2.2\: m = 2.2 \times 39.37 = 86.614\: in$

Question (ii)

Substitute the found value of D from part (i) into the given formula for D and solve for S:

$\sf \implies D=\dfrac{S}{0.55}$

$\sf \implies 86.614=\dfrac{S}{0.55}$

$\implies \sf S = 86.614 \times 0.55$

$\implies \sf S = 47.6377\: in$

Question (iii)

Given:

• E = 1.1 m
• D = 2.2 m

Substitute the given values into the given formula for M and solve for M:

$\implies \sf M = E + D \times 0.22$

$\implies \sf M = 1.1 + 2.2 \times 0.22$

$\implies \sf M = 1.1 +0.484$

$\implies \sf M = 1.584 \: m$

Part (b)

Given:

• S = 50 in
• Ratio of width to height = 16 : 9

Pythagoras Theorem

$a^2+b^2=c^2$

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:

$\implies \sf 16^2+9^2=S^2$

$\implies \sf 256+81=S^2$

$\implies \sf S^2=337$

$\implies \sf S=\sqrt{337}$

Therefore, the ratio of width to height to diagonal S of the TV is:

$\sf w:h:S=16 : 9 : \sqrt{337}$

If S = 50 in then:

$\implies \sf w:50=16:\sqrt{337}$

$\implies \sf \dfrac{w}{50}=\dfrac{16}{\sqrt{337}}$

$\implies \sf w=\dfrac{16}{\sqrt{337}} \times 50$

$\implies \sf width=43.57877686..\:in$

To convert to meters, divide the width in inches by 39.37:

$\implies \sf width=\dfrac{43.57877686}{39.37}=1.106903146\:m$

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:

$\implies \textsf{Top of TV} \sf = 1.584 + \dfrac{62.2 \div 100}{2}$

$\implies \textsf{Top of TV} \sf = 1.895\:m$

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:

$\implies \sf 1.895-1.1=0.795\:m$

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

$\sf \tan(\theta)=\dfrac{O}{A}$

where:

• $\theta$ is the angle
• O is the side opposite the angle
• A is the side adjacent the angle

Given:

• $\theta$ = 35°
• O = 0.795 m
• A = D m

Substitute the values into the formula and solve for D:

$\implies \sf \tan(35^{\circ})=\dfrac{0.795}{D}$

$\implies \sf D=\dfrac{0.795}{\tan(35^{\circ})}$

$\implies \sf D=1.135377665\:m$

Therefore, the shortest distance that the sofa should be placed from the wall is 1.135 m (3 d.p.).