2. What is the related angle for each of the following: [2 marks]

a. 50°
b. 120°


3. Jillian has to replace the two supporting guy wires for a hydro pole. She measures the distance between the bases of the wires to be 7 m, and their angles of inclination to be 40° and 30°.

This link is to the photo

https://imgur.com/a/NsoefWT

(if it cannot be loaded, this image just has a triangle drawn out with dimensions given above, right angle in bottom left corner, top left being the "top" of the hydro pole, and two lines going down to the right with the 40 and 30, with the 7m being drawn inbetween the 40 and 30)

a) Determine the total length of guy wire that needs to be replaced, to the nearest metre. [3 marks]
b ) Determine the height of the pole, to the nearest metre. [1 mark]

4. a) Calculate the value(s) of theta (the angle) and its supplementary angle. [2 marks]

sinθ = 1/4


b ) What is the difference between the following two questions: [1 mark]
i. Calculate the angle(s).
ii. Calculate the angle given that θ>90°.



This is trigonometry

would really appreciate help
really struggling with this assignment

This post was edited by RollOneUp on Jan 3 2023 04:38pm

or

if you guys dont mind helping, would be forever grateful

2. What is the related angle for each of the following: [2 marks]

a. 50°
b. 120°


Been teaching this for 7 years and never heard of related angles. I am guessing it is reference angles and angles it is a cultural difference (like UK vs US)

Reference angles are the acute angles formed with the horizontal axis In standard position.

There’s a bunch of formulas for these. I don’t recommend memorizing them. Just draw each angle on its own x-y coordinate grid with 0 degrees being the positive x axis.

When you draw 50 degrees you should ask yourself how far (smallest) is this away from the x axis in either side. It’s 50 degrees away. So it’s reference is 50 degrees.

When you draw 120 degrees you should ask yourself how far (smallest) is this away from the x axis on either side. It’s 60 degrees away from the left side of the x axis. So it’s reference is 60 degrees.

Here is a link that draws the pictures for you.
https://www.omnicalculator.com/math/reference-angle


I’ll look at the next between games of smite.

I can’t see the image. I drew two right triangles.

One has
Hypotenuse(guy wire)- r1
Base- x (distance from first wire to pole)
Height (pole) - h

Second has
Hypotenuse (guy wire)- r2
Base- x+7 (distance from second wire to pole)
Height (pole) - h

Set up a system

Tan 40 = h/x

Tan 30 = h/(x+7)

Solve both for h

x tan 40 = h

(x+7)tan30 = h

Both equal h so we can set the left expressions equal to each other

xtan40 = (x+7)tan30

Distribute right

x tan40 = x tan30 + 7 tan 30


Rewrite to have terms with variable on same side

x tan 40 - x tan 30 = 7 tan 30

Factor out x

x ( tan 40 - tan 30) = 7 tan 30

Solve for x

x = (7tan30)/(tan40-tan30)

We know x. We can find h


h = x tan 40. This was one of our equations

h = ((7tan30)/(tan40-tan30)) (tan40)

I’ll stop here for a sec in a match

h = ((7tan30)/(tan40-tan30)) (tan40) = 12.955832 to nearest meter =13m

We can now find the length of guy wire needed that would be r1+r2

Sin40=h/r1

r1 sin 40= h

r1 = h/sin40

r1 = ( ((7tan30)/(tan40-tan30)) (tan40) ) / sin40 = 20.155696691


Sin30=h/r2

r2 sin 30= h

r2 =h/sin30

r2= ( ((7tan30)/(tan40-tan30)) (tan40) ) / sin30= 25.911664195

r1+r2=20.155696691+25.911664195=46.0673608862

Your question says to the nearest meter, but This should be rounded up to 47 since 46 would be short. This would depend on whether your instructor is pushing the whole real world thing or not.

regarding 2:
the concept of 'related angles' is new for me as well. i am from germany, so i am not that familiar with the english or american terms. google didnt give me a decisive definition either...

regarding 3:
We can see two triangles with a right angle, one is embedded inside the other. With respect to the given angles, the bottom side is the adjacent while the pole represents the opposite. If we label the length of the pole p and the length of the bottom side of the smaller triangle d, we get two equations for two variables:
tan(40deg) = p/d
tan(30deg) = p/(d+7)
Solving this set of equations yields p=12.9558 (i.e.13, this is already the answer to a) ) and d=15.4402. With these results, we can calculate the hypotenuses h1 and h2 in both triangles by employing either sine or cosine:
small: sin(40deg) = p/h1, so h1 = 20.1556
large: sin(30deg) = p/h2, so h2 = 25.9116
The total length of wire is thus
w = h1+h2 = 46.0672 (i.e. 46 or 47 depending on if you think while rounding or if you dont, i guess...)

regarding 4:
The first step is to employ the inverse sine to solve for principal value theta1:
theta1 = arcsin(1/4) = 14.48deg
However, since the sine function is periodic, there are infinite values of theta that fulfill this equation. Sticking to elementary geometry, we may just calculate the ones in a range of 0-180deg. There is only one other possibility, given by theta2 = 180deg-arcsin(1/4) = 165.52deg
Google tells me that 'supplementary angles' add up to 180deg, so theta1 and theta2 are already each other's supplementary angle by construction.
Task b again refers to this concept: 'Calculate the angle(s)' means that there may be multiple solutions (in these easy cases: 2) where as the restriction 'theta > 90deg' gives only one solution.

thank you both will get this done when off work later

State the three primary trigonometric ratios of the angle, , in standard position with terminal arm passing through P(-3, 5).

if you can please help with this answer, im in an exam now and have no idea its not in any of the reviews , etc

its sine, cosine, and tangent but the second part i dont understand ?

This post was edited by RollOneUp on Jan 11 2023 06:16pm

A plane is approaching a 7500 m runway. The angles of depression to the ends of a runway are 9 and 16 degrees. How far is the plane from each end of the runway? Show all work for full marks.


also this one
have about an hour left on the exam timer, would really appreciate this

You exam is probably over but it sounds like just SOH CAH TOA.

Draw a right triangle using the points (-3,5) and base of the x axis.

You’ll notice the base has “length” -3

Height 5

And you can find the hypotenuse by using the Pythagorean theorem.

(-3)^2 + (5)^2 = c^2.

9 + 25 = c^2

34 = c^2

Sqrt 34 = c. Now label your triangle



Sin theta = 5/ sqrt 34. Most instructors will want you to rationalize your denominator. This means no sqrt in the bottom of the fraction. Multiple your fraction by sqrt34 / sqrt34

Sin theta = (5 sqrt 34) / 34

Cos theta = -3/ sqrt 34
= (-3 sqrt 34)/ 34

Tan theta = 5/-3