Applying trigonometric ratios, the missing segments lengths in the image attached below are: <em>a = 8√3; b = 12.</em>
<h3>The Trigonometric Ratios</h3>
- SOH is sin ∅ = opp/hyp
- CAH is cos ∅ = adj/hyp
- TOA is tan ∅ = opp/adj
- SOHCAHTOA is used to solve right triangles.
Given the right triangle in the mage attached below, we would find the missing sides as follows:
<em>Find a:</em>
Reference angle (∅) = 30°
opposite = 4√3
hypotenuse = a
sin 30 = (4√3)/a
a = (4√3)/sin 30
a = (4√3)/(1/2) (sin 30 = 1/2)
a = (4√3)×2
a = 8√3
<em>Find b:</em>
Reference angle (∅) = 60°
opposite = b
adjacent = 4√3
tan 60 = b/(4√3)
b = tan 60 × 4√3
b = √3 × 4√3
b = 12
Therefore, applying trigonometric ratios, the missing segments lengths in the image attached below are: <em>a = 8√3; b = 12.</em>
Learn more about trigonometric ratios on:
brainly.com/question/4326804
Answer:
area of BMDN 8 ft²
Step-by-step explanation:
In BMDN the diagonals are BD and MN.
MN is 12/3 = 4 ft long, and BD = 4 ft.
A rhombus with equal diagonals is a square, so BMDN is a square.
In the right triangle BDN, the diagonal BD is the hypotenuse, then:
DN² + NB² = BD²
but DN = NB, then:
2*DN² = 16 ft²
DN² = 8 ft²
The area of BMDN is computed as one of its sides squared. Then, DN² is its area
Let slant height be s
Height = 38yd. = h
Side = 42yd. = a
S² = h² + (a/2)²
= 38² + 21²
= 1444 + 441
= 1885
S = √1885
S = 43.42yd.
Lateral area = 1/2 × 4a × s
= 2 × 42 × 43.42
= 3646.99
= 3647 yd.²
Total distance traveled = 4 + 6 = 10 miles
Total time = 15 + 25 = 40 minutes
Speed = 10 miles ÷ 40 minutes = 0.25 mph