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trigonometry

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There isn't a question, but assuming it's similar to the last, how would you go about getting extra angles (and side) and area.

I have to find aBc and
given that the bAc angle is acute what it is
and the area

Reply 21

mqb2766

Original post by louise18152

I have to find aBc and
given that the bAc angle is acute what it is
and the area

So ...
A useful hint is to try and draw sketch in proportion.
So 25 degree angle is smaller, sides should be in proportion and the stated angle is acute.
Helps reduce errors.

(edited 3 years ago)

Reply 22

louise18152

Original post by mqb2766

So ...

i did sin25/12 =sinc/19 which equals 0.669
then sin-1 (0.669)=42
but idk for the the second part

Reply 23

mqb2766

Original post by louise18152

i did sin25/12 =sinc/19 which equals 0.669
then sin-1 (0.669)=42
but idk for the the second part

Is the angle acute?
If not try and redraw/sketch the triangle and think how the sin rule gives two possible answers.

Reply 24

louise18152

Original post by mqb2766

Is the angle acute?
If not try and redraw/sketch the triangle and think how the sin rule gives two possible answers.

the ambiguous rule
so would I do 180 -42 - 25=113 then the oppisite

Reply 25

mqb2766

Original post by louise18152

the ambiguous rule
so would I do 180 -42 - 25=113 then the oppisite

But 113 is not acute. So the sin rule is (slightly) wrong.
Have you sketched the triangle a bit more accurately? It helps get your intuition right.

Reply 26

louise18152

Original post by mqb2766

But 113 is not acute. So the sin rule is (slightly) wrong.
Have you sketched the triangle a bit more accurately? It helps get your intuition right.

i don't know how else I'm meant to sketch it

Reply 27

mqb2766

Original post by louise18152

i don't know how else I'm meant to sketch it

Your 25 degrees is closer to 90. One side should be almost twice the other. You want the last angle to be acute.

Original post by vanshika20

Hope this helps :smile:

If the picture is unclear, or if you have any doubts, feel free to ask!

Can you pls delete and read the sticky at the top of the forum about not providing solutions.

Reply 30

vanshika20

Original post by mqb2766

Can you pls delete and read the sticky at the top of the forum about not providing solutions.

I am sorry, I don't get what you mean.

Reply 31

mqb2766

Original post by vanshika20

I am sorry, I don't get what you mean.

This
https://www.thestudentroom.co.uk/showthread.php?t=4919248
The aim is to help students, not do their work.

Reply 32

louise18152

Original post by vanshika20

I am sorry, I don't get what y

Original post by mqb2766

Can you pls delete and read the sticky at the top of the forum about not providing solutions.

im still stuck what do i do for the acute angle

Reply 33

mqb2766

Original post by louise18152

Better, so how do you get an obtuse C?
It's still the sin rule, but what do you get when you take asin()?

Reply 34

mqb2766

Original post by vanshika20

Could you please look at my working? What do you not understand in that??

That question has been solved.
Pls delete.

Reply 35

vanshika20

Original post by mqb2766

This
https://www.thestudentroom.co.uk/showthread.php?t=4919248
The aim is to help students, not do their work.

oh okayyy. I am so sorry, my bad:/

Reply 36

louise18152

Original post by mqb2766

Better, so how do you get an obtuse C?
It's still the sin rule, but what do you get when you take asin()?

I'm still confused

Reply 37

mqb2766

Original post by louise18152

I'm still confused

Ok.You must have covered trig for obtuse angles? Cast diagram or ...
Your previous sin rule is correct. But when you take asin, you want the obtuse solution rather than the acute.

Note
sin(160) = sin(20)
sin(130) = sin(50)
sin(100) = sin(80)
....

Reply 38

louise18152

I understand that but I dont know what sin is