Think of it as a normal linear equation first. Let's find the slope.
m = rise/run = (3-1)/(0-1) = -2
We know the slope is negative now, so we can immediately get rid of the first two answers. Now, we know that the solutions must be under the line itself, so we can try figuring it out by testing some points. Let's use (0,0).
Is 0 </> 0+3? Since it's <, then we know the last answer is correct (y < -2x + 3).
Answer:
tan(2u)=[4sqrt(21)]/[17]
Step-by-step explanation:
Let u=arcsin(0.4)
tan(2u)=sin(2u)/cos(2u)
tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]
If u=arcsin(0.4), then sin(u)=0.4
By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.
This also implies cos(u)=sqrt(0.84) since cosine is positive.
Plug in values:
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]
tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]
tan(2u)=[(40)(sqrt(0.84)]/[34]
tan(2u)=[(20)(sqrt(0.84)]/[17]
Note:
0.84=0.04(21)
So the principal square root of 0.04 is 0.2
Sqrt(0.84)=0.2sqrt(21).
tan(2u)=[(20)(0.2)(sqrt(21)]/[17]
tan(2u)=[(20)(2)sqrt(21)]/[170]
tan(2u)=[(2)(2)sqrt(21)]/[17]
tan(2u)=[4sqrt(21)]/[17]
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tinyurl.com/wpazsebu
Given :
A sailor is 30 m above the water in the crow's nest on a sailboat.
The sailor encounters an orca surface at an angle of depression of 15 degrees.
The crows nest is 20 m horizontally from the bow (front) of the boat.
To Find :
How far in front of the boat is the orca.
Solution :
Let, distance of boat front from the crow's nest is x.
So,
Hence, this is the required solution.
