# Thread: Trig - solving triangles

1. ## Trig - solving triangles

I have a triangle, two side lengths are known and one angle, then a right angle triangle is added to the triangle, please see diagram below.

I am asked to find the length of AD, I used the sine rule to find it.

I am asked to calculate the angle ABD, I used sin theta = opp / hyp.

I am asked to calculate angle ABC, I subtracted 180 - theta = angle ABC

I am asked to find the length DC, given I worked out AD and knew AC I figured Pythagoras could be used.

This is where the problems begin!

I was asked to find the length DB, now I have calculated the length AD and DC, and the hypotenue is known at 8cm, however after using Pythagoras to work the length out DB which was calculated at 1.23cm, I found that DC - DB = 2.25cm and not 1.23cm using Pythagoras,

So my question is;

Looking at the diagram I have a triangle with a right angle triangle added to it, there is two hypotenuse 3 and 8, but I think triangle ADB cannot have pythagoras used to solve the lengths of the sides?

Thanks

David

2. ## re: Trig - solving triangles

For the sake of easy checking, what answers did you get when you calculated the length of AD, the angle ABD, angle ABC, and the length DC?

3. ## re: Trig - solving triangles

Originally Posted by Quacky
For the sake of easy checking, what answers did you get when you calculated the length of AD, the angle ABD, angle ABC, and the length DC?
ABD = 66 degrees
ABC = 114 degrees
DC = 7.52cm

Thanks

David

4. ## re: Trig - solving triangles

Originally Posted by David Green
I have a triangle, two side lengths are known and one angle, then a right angle triangle is added to the triangle, please see diagram below.

I am asked to find the length of AD, I used the sine rule to find it.

I am asked to calculate the angle ABD, I used sin theta = opp / hyp.

I am asked to calculate angle ABC, I subtracted 180 - theta = angle ABC

I am asked to find the length DC, given I worked out AD and knew AC I figured Pythagoras could be used.

This is where the problems begin!

I was asked to find the length DB, now I have calculated the length AD and DC, and the hypotenue is known at 8cm, however after using Pythagoras to work the length out DB which was calculated at 1.23cm, I found that DC - DB = 2.25cm and not 1.23cm using Pythagoras,

So my question is;

Looking at the diagram I have a triangle with a right angle triangle added to it, there is two hypotenuse 3 and 8, but I think triangle ADB cannot have pythagoras used to solve the lengths of the sides?

Thanks

David
Originally Posted by David Green
ABD = 66 degrees
ABC = 114 degrees
DC = 7.52cm

Thanks

David
I wouldn't use the sine rule here because we have the ambiguous case.

I would use the fact that $\displaystyle \sin{20}=\frac{AD}{8}$
Which gives $\displaystyle AD=2.74$ which is what you have (which is correct).

I agree with all solutions so far. I also agree that $\displaystyle DB=1.22$

$\displaystyle DC-DB=BC=6.3$ going by what we have so far.

We can confirm that $\displaystyle BC=6.3$ using the cosine rule:

$\displaystyle (BC)^2=8^2+3^2-2(8)(3)cos(46)$
$\displaystyle =64+9-48cos(46)$
which comes out at $\displaystyle BC=6.3$

$\displaystyle DC-BC=1.22$, as we'd expect.

5. ## re: Trig - solving triangles

Originally Posted by Quacky
I wouldn't use the sine rule here because we have the ambiguous case.

I would use the fact that $\displaystyle \sin{20}=\frac{AD}{8}$
Which gives $\displaystyle AD=2.74$ which is what you have (which is correct).

I agree with all solutions so far. I also agree that $\displaystyle DB=1.22$

$\displaystyle DC-DB=BC=6.3$ going by what we have so far.

We can confirm that $\displaystyle BC=6.3$ using the cosine rule:

$\displaystyle (BC)^2=8^2+3^2-2(8)(3)cos(46)$
$\displaystyle =64+9-48cos(46)$
which comes out at $\displaystyle BC=6.3$

$\displaystyle DC-BC=1.22$, as we'd expect.
Thank you for taking the time to work through the trig for me, I can see now that the mistake a made was using the 20 degrees where you used 46 degress, which gives a slightly longer length BC than using angle 20 degrees.

so using pythagoras then resulting in a answer of 1.23 against your 1.22 is this completely wrong or a rounding error?

Thanks

David

6. ## re: Trig - solving triangles

Originally Posted by David Green
I have a triangle, two side lengths are known and one angle, then a right angle triangle is added to the triangle, please see diagram below.

I am asked to find the length of AD, I used the sine rule to find it.

I am asked to calculate the angle ABD, I used sin theta = opp / hyp.

I am asked to calculate angle ABC, I subtracted 180 - theta = angle ABC

I am asked to find the length DC, given I worked out AD and knew AC I figured Pythagoras could be used.

This is where the problems begin!

I was asked to find the length DB, now I have calculated the length AD and DC, and the hypotenue is known at 8cm, however after using Pythagoras to work the length out DB which was calculated at 1.23cm, I found that DC - DB = 2.25cm and not 1.23cm using Pythagoras,

So my question is;

Looking at the diagram I have a triangle with a right angle triangle added to it, there is two hypotenuse 3 and 8, but I think triangle ADB cannot have pythagoras used to solve the lengths of the sides?