Hi new here.
ABCD is a paralleologram in which the lengths of AB and AD are 5 cm and 12 cm respectively and angle ABC is 50 degrees.
calculate
a) length of diagonal AC , correct to 2 dp.
b) size of angle BAC
Hi new here.
ABCD is a paralleologram in which the lengths of AB and AD are 5 cm and 12 cm respectively and angle ABC is 50 degrees.
calculate
a) length of diagonal AC , correct to 2 dp.
b) size of angle BAC
Hello thcbender
Welcome to Math Help Forum!The length of $\displaystyle BC$ is also $\displaystyle 12$ cm. So use the Cosine Rule on $\displaystyle \triangle ABC$:
$\displaystyle AC^2 = AB^2+BC^2-2AB.BC\cos \angle ABC$Once you have the length of $\displaystyle AC$, use the Sine Rule on the triangle to calculate $\displaystyle \angle BAC$.
Can you complete this now?
Grandad
Hello,
the best way to approach problems of this type is to draw a rough sketch and use pythogoras theorem multiple times as required, remove the overlapping parts added more than once appropriate number of times.
This is a general approach suitable for many such problems. The exact answer is as grand dad has stated above.
You can try deriving it as you will remember it better that way rather than to just look for the answer directly! Also remember sine = op/hyp , cos = adj/hyp of the angle considered
I hope I did this right:
I've drawn up a diagram:
For the first diagonal question, it is really easy.
See the right angled triangle I made? Yea just use pythagoras.
First we need to work out x:
$\displaystyle \cos 40 = \frac {x}{12} $
$\displaystyle x= 9.19253 $ (i'm skipping the workings but you should write them)
now find y:
$\displaystyle \sin 40 = \frac {y}{12} $
$\displaystyle y= 7.71345 $
Now add y onto 5... we get 12.7135
Set up pythag
Let diagonal be t. I know it's random :P
$\displaystyle t = \sqrt (12.7135^2 + 9.19253^2) $
Spoiler:
For the next question, if you draw a diagonal through A to C. There should be an angle that is part of the right angled triangle I made. Get that angle and subtract 40.
Lets do it:
$\displaystyle \sin\theta = \frac {12.7135}{15.6887} $
You have to show working out but I'll just give you the answer for now (i'm fast at typing normally but not good with latex :P)
$\displaystyle = 54.1312 $
Now minus 40
we get
$\displaystyle 14.1312 $
Alternatively we could have used the sine rule but this rules out any hard stuff
Hello everyone
Provided you know the Cosine Rule and Sine Rule for any triangle, you'll find the quickest way is to use the method I outlined.
In jgv115's diagram, $\displaystyle \angle ABC = 130^o$. The question stated $\displaystyle \angle ABC = 50^o$.
Grandad