if you do sin 30 = opp/ 16 , opp = 8cm Then via pythag, you get 8 root 3 as the height
However, if you do sohcohtoa twice, both sides become 8cm.
When I checked pythag, 8 root 3 worked so I stuck with that
You cannot use Pythagoras as you don't have the length of the side AB but if you did work it out, then AB= 16cos(30)=8root3 but this is the length of the side Ab and not the height .
Because the other triangle has two equal sides, one of which is the height of the triangle we have just found out, then the height of the right hand side triangle would be h=root(82-62)= 2root7 . Therefore the area of the right hand side triangle would be 1/2*(6)(2root7)=3root7
You cannot use Pythagoras as you don't have the length of the side AB but if you did work it out, then AB= 16cos(30)=8root3 but this is the length of the side Ab and not the height .
Because the other triangle has two equal sides, one of which is the height of the triangle we have just found out, then the height of the right hand side triangle would be h=root(82-62)= 2root7 . Therefore the area of the right hand side triangle would be 1/2*(6)(2root7)=3root7
There are so many way you can do this questions, so bizarre.
I found length AB by doing 16cos60 = 8 Then found length BC using pythag = 8 root 3
So the area of the right angled triangle (left triangle) would be 0.5 x 8 x 8 root 3 = 32 root 3
The right hand side triangle is isosceles so you would need to split that in half, to get two right angles, then find the height via pythag again. then you can use the 1/2 x base x height formula
I am intrigued to see what the answer is for this...
There are so many way you can do this questions, so bizarre.
I found length AB by doing 16cos60 = 8 Then found length BC using pythag = 8 root 3
So the area of the right angled triangle (left triangle) would be 0.5 x 8 x 8 root 3 = 32 root 3
The right hand side triangle is isosceles so you would need to split that in half, to get two right angles, then find the height via pythag again. then you can use the 1/2 x base x height formula
I am intrigued to see what the answer is for this...
You cannot use Pythagoras as you don't have the length of the side AB but if you did work it out, then AB= 16cos(30)=8root3 but this is the length of the side Ab and not the height .
Because the other triangle has two equal sides, one of which is the height of the triangle we have just found out, then the height of the right hand side triangle would be h=root(82-62)= 2root7 . Therefore the area of the right hand side triangle would be 1/2*(6)(2root7)=3root7
I think it depends on what you are using as your angle, I used 60 degree angle at C to work out AB using 16cos60 = 8 (base) and height was found via pythag
Oh right yeah then the 8rt3 of abc isn't relevant if she needs to find area of bcd but you don't need to half it do you?
You could work that out to find the height of the triangle ABC using the Pythagoras but that would involve another step in calculation which is not really preferred.
I think it depends on what you are using as your angle, I used 60 degree angle at C to work out AB using 16cos60 = 8 (base) and height was found via pythag
I think it depends on what you are using as your angle, I used 60 degree angle at C to work out AB using 16cos60 = 8 (base) and height was found via pythag
It still gives you 8 as the height though as a and o would be different depending on the angle