Answer:
212m
Step-by-step explanation:
The set up will be equivalent to a right angled triangle where the height is the opposite side facing the 45° angle directly. The length of the rope will be the slant side which is the hypotenuse.
Using the SOH, CAH, TOA trigonometry identity to solve for the length of the rope;
Since we have the angle theta = 45° and opposite = 150m
According to SOH;
Sin theta = opposite/hypotenuse.
Sin45° = 150/hyp
hyp = 150/sin45°
hyp = 150/(1/√2)
hyp = 150×√2
hyp = 150√2 m
hyp = 212.13m
Hence the length of the rope for the kite sail, in order to pull the ship at an angle of 45° and be at a vertical height of 150 m is approximately 212m
C=90°, A=75°, b=AC=19, x=AB
Without a figure, we see AC is adjacent to angle A, so
cos A = AC/AB = b/x
x = b/cos A = 10 / cos 75° ≈ 38.637
Answer: 38.6
Answer:
Options (3) and (6)
Step-by-step explanation:
ΔABC is a dilated using a scale factor of
to produce image triangle ΔA'B'C'.
Since, dilation is a rigid transformation,
Angles of both the triangles will be unchanged or congruent.
m∠A = m∠A' and m∠B = m∠B'
Since, sides of ΔA'B'C' =
of the sides of ΔABC
Area of ΔA'B'C' = 
Area of ΔABC > Area of ΔA'B'C'
Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.
ΔABC ~ ΔA'B'C'
Therefore, Option (3) and Option (6) are the correct options.
To find the answer, you should divide 405 by 50 to find the mass of one coin. The formula should look like this:

= 8.1
The exact mass is 8.1 grams, but because you want an estimate, the answer should be
About 8 grams for the mass of 1 one-dollar coin