SURFACE AREAS AND VOLUMES (10Th Grade)
2 answers:
You might be interested in
a. See attachment for the labelled diagram
b. Using the sine ratio, distance from the hot air to pond A is 5,038.9 m, while distance from the hot air to pond B is 6,287.1 m.
c. Distance between the two ponds is 1,263.5 m.
<h3>What is the Sine and Tangent Ratios?</h3>Sine ratio, is: sin ∅ = opposite side/hypotenuse length
Tangent ratio is: tan ∅ = opposite side/adjacent length.
a. The diagram with the appropriate labels is shown in the image attached below.
b. Use the sine ratio to find the distance from the hot air to pond A (CA) and to pond B (CB):
CA = hypotenuse
∅ = 10°
Opposite = 875 m
sin 10 = 875/CA
CA = 875/sin 10
CA ≈ 5,038.9 m (distance from the hot air to pond A)
CB = hypotenuse
∅ = 8°
Opposite = 875 m
sin 8 = 875/CB
CB = 875/sin 8
CB ≈ 6,287.1 m (distance from the hot air to pond B)
c. Distance between the two ponds, BA = BD + DA.
Apply the tangent ratio to find BD and DA
tan 8 = 875/BD
BD = 875/tan 8
BD = 6,225.9 m
tan 10 = 875/DA
DA = 875/tan 10
DA = 4,962.4 m
Distance between the two ponds = 6,225.9 - 4,962.4 = 1,263.5 m.
Learn more about the sine and tangent ratios on:
#SPJ1
