helppppppp, difficult maths question

I found that bit wierd because:

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

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

Isn't the height 8 and the base 8rt3

@rede121

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16sin30 =8 height
16cos30 = 8rt3 base

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?

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

Isn't the area of bcd 12rt7?

Oh sorry! It should be 6root7 as Area= 1/2(6)(2root7)

But there are 2 triangles in bcd so it's like the area of square

That's totally correct, my mistake! I'm glad I don't need to go back to GCSE

It still gives you 8 as the height though as a and o would be different depending on the angle

Ok, I typed it up, does this make more sense? This question is going to be the bane of my life

Haha yeah on the first one you drew you wrote the height as 8rt3 that's why I was confused