HELP!!!! TRIGINOMETRY QUESTION!!!!!! Help!!!!?

Question

Here is the question. I really need help! It's pretty basic I know but I'm just 14 so I'm not a maths genius or anything.

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Answer 1

Tangent of 24 = DB over 20
to get width DB

and tangent of 14 = AB over 20
to get width AB

then its just subtracting and adding the values to get the answers

Answer 2

Not sure if you've drawn it correctly, but I think you mean the distance AB is 20 m.

Looking at the Triangle(ABC),

AB = AC sin(14deg), and
BC = AC cos(14deg)

So compute AC from the first equation, then compute BC using the result.

Looking at the Triangle(BCD),

BD = BC tan(10 + 14 deg)

The width of the spotlight, AD is just the difference between BD and AB.

Last of all, if your spotlight points straight down, it will shine 5 deg to the left of B on the floor and 5 deg to the right of B (Make another sketch to guide yourself). The distance from B on either one side will be BC tan (5 degree). Since you have left + right, you've got to double this amount to get the width of the spotlight there.

Answer 3

it particularly is merely the hardship-loose Definitons: the form of a ideal triangle is thoroughly desperate, as much as similarity, by capacity of the fee of the two of the different 2 angles. this means that once one among the different angles is often used, the ratios of the various components are continually the comparable inspite of the dimensions of the triangle. those ratios are traditionally defined by capacity of right here trigonometric applications of the regularly occurring attitude: The sine function (sin), defined because of the fact the ratio of the leg opposite the attitude to the hypotenuse. The cosine function (cos), defined because of the fact the ratio of the adjoining leg to the hypotenuse. The tangent function (tan), defined because of the fact the ratio of the different leg to the adjoining leg. The adjoining leg is the fringe of the attitude that's no longer the hypotenuse. The hypotenuse is the side opposite to the ninety degree attitude in a ideal triangle; this is the longest fringe of the triangle. The words perpendicular and base are from time to time used for the different and adjoining components respectively. The reciprocals of those applications are named the cosecant (csc or cosec), secant (sec) and cotangent (cot), respectively. The inverse applications are referred to as the arcsine, arccosine, and arctangent, respectively. There are arithmetic kin between those applications, that are regularly occurring as trigonometric identities. With those applications you may actually answer very almost all questions approximately arbitrary triangles by capacity of utilising the regulation of sines and the regulation of cosines. those regulations could nicely be utilized to compute the relax angles and components of any triangle as quickly as 2 components and an attitude or 2 angles and a side or 3 components are regularly occurring. those regulations are useful in all branches of geometry when you consider that each and every polygon could be defined as a finite blend of triangles. i'm hoping this could help!! ^_^

Answer 4

AB / BC = tan 14
AB / 20 = 0.2493
AB = 4.986

BD / BC = tan 24
BD = 8.904

AD = BD - AB = 3.918 m

When the beam is directly below C, half of the beam is towards either side of BC. The angle C on either side is therefore 5 degree

Let MN mark the beam.

Therefore
MB / BC = tan 5
MB = 1.75

MN = 2MB = 3.5 m

It would be clear if you draw a separate figure for the 2nd part.

Hope this helps.

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Answer 5

AB = 20tan14°
DB = 20tan24°
The LENGTH of the beam
AD = 20(tan24° - tan14°) ≈ 3.9180 m or 3.92 m
Let E be the point where the center of the beam hits the floor.
CE = 20/cos19° ≈ 21.15241 m
The WIDTH of the beam
W = 2(21.15241)tan5° ≈ 3.701193 m or 3.70 m

The width of the beam at B is
2(20)tan5° ≈ 3.499547 m or 3.500 m

Answer 6

distance from pt.A to pt.B is
AB=20meters x (tan 14 degrees)
AB= 4.986 m

distance from pt.D to pt. B is
DB=20 meters x (tan (10+14) degrees)
DB=8.9046


distance from pt.D to pt.A is
DA=DB - AB
DA = 8.9046 - 4.9866
DA = you do the work here

Answer 7

formula:
tan(angle)=opossite/adjacent

tan(14)=AB/20 >> AB = tan(14)*20 =4.99

tan(10+14)=tan(24)=DB/20 >> DB=tan(24)*20=8.90

AD=DB-AB=8.90-4.99=3.91

About the second question idk, my english is not very good yet to understand the question

Answer 8

....refer to a trigonometry book!