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3Tvwo §6.4 / 3Thavo §6.5 Calculating in right-angled triangles.
§6.4 / §6.5 Calculating in right-angled triangles
Choose the right Formula-in-letters, that tells you how to calculate the
SINE, COSINE and TANGENT, in the next slide.
1 / 21
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Slide 1:
Slide
Wiskunde
Middelbare school
havo, vwo
Leerjaar 3
This lesson contains
21 slides
, with
interactive quizzes
and
text slides
.
Lesson duration is:
50 min
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Items in this lesson
§6.4 / §6.5 Calculating in right-angled triangles
Choose the right Formula-in-letters, that tells you how to calculate the
SINE, COSINE and TANGENT, in the next slide.
Slide 1 - Slide
vwo §6.4 / havo §6.5
Calculating in right-angled triangles
Because it's been a while since you worked on this Chapter -holiday!- we will look back shortly on the
content you learned up to now!
VWO classes: STUDY SQUARERS, feel free to go to the STUDY SQUARE, doing § 6.4 on your own.
Slide 2 - Slide
The correct letterformula is:
A
SAH COA TAH
B
SIN COS TOA
C
COH TAH SIN
D
SOH CAH TOA
Slide 3 - Quiz
Answer: SOH CAH TOA
meaning:
Sine = Opposite shorter side / Hypotenuse
Cosine = Adjacent shorter side / Hypotenuse
Tangent = Opposite shorter side / Adjacent shorter side
Slide 4 - Slide
Slide 5 - Slide
§6.4/§6.5 Calculating in right-angled triangles
You now know the:
+ sine, cosine and tangent
+ but when do you use which of these?
+ that is shown in the next slides!
+ FIRST WE DO 21 TOGETHER!
Slide 6 - Slide
Slide 7 - Slide
Slide 8 - Slide
Slide 9 - Slide
Homework time.
+ Do §6.4 vwo / §6.5 havo Calculating in right-angled triangles
+ Later a new technique will be explained, that you need to do the last 3 exercises!
Slide 10 - Slide
Explanation for the last 3 exercises of the paragraph.
Slide 11 - Slide
If you want to calculate length PQ,
then what is the problem?
Key in your answer in the next slide!
Slide 12 - Slide
What's the pro, bro (sis)?
So: describe the problem!
Slide 13 - Mind map
Answer:
The big problem is:
there is no right angle to be seen?!
Pythagoras only works in right-angled triangles!
The method to solve this problem is shown in the next slide.
Slide 14 - Slide
Slide 15 - Slide
Slide 16 - Slide
Now there are even two right-angled triangles!
REMEMBER: WE ARE AFTER: length PQ!!
1. draw auxiliary line (=helplijn) RS.
2. calculate RS with sine angle P
3. calculate PS with cosine angle P (or tangent angle P)
4. calculate QS with tangent angle Q
5. add PS to QS to get length PQ
Slide 17 - Slide
helping line = auxiliary line
Slide 18 - Slide
Homework time again.
In the next slide an Oldie-but-Goldie...
Slide 19 - Slide
Slide 20 - Slide
Slide 21 - Slide
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