the AC is not a straight line by the way. Just afraid my drawings are not clear(Wink)

Printable View

- Jul 12th 2008, 09:24 PMhelloyinghelp me solve this considered easy maths question
the AC is not a straight line by the way. Just afraid my drawings are not clear(Wink)

- Jul 12th 2008, 09:47 PMJhevon
see the modified diagram below. Note that AB = E + F, where E and F are the bases of the two right-triangles we've constructed. we can find their lengths using trig ratios:

$\displaystyle \sin 40^o = \frac E8$

$\displaystyle \cos 25^o = \frac F9$

EDIT: i forgot about finding the angle CAD. Note that you can find the length of BC in the same way we found AB. once that is found, we can use Pythagoras' theorem to find the length AC. Thus we will know the lengths of all the sides of triangle ACD, and we can use the law of cosines to find any angle in that triangle we want. - Jul 12th 2008, 09:49 PMhelloying
oh i see thanks for ur help. u are so smart.

- Jul 12th 2008, 09:53 PMJhevon
- Jul 12th 2008, 09:57 PMticbol
**Ooopss....**

I am very slow in typing. Let me post this anyway.

Let AB be horizontal. So, BC is vertical.

From D, drop a vertical line seggment to AB. Call the intersection as point E.

From D also, draw a horizontal line segment to touch BC. Call the intersection as point F.

----------------------------

1) To find lenght of BC:

BC = BF +FC

Since BF = DE, then, BC = DE +FC ----**

In right triangle AED,

sin(25deg) = DE / 9

so, DE = 9sin(25deg) = 3.80356 units.

In right triangle CFD,

cos(40deg) = FC / 8

So, FC = 8cos(40deg) = 6.12836 units

therefore, BC = 3.80356 +6.12836 = 9.93192 units long. ------answer.

------------------------

To find angle CAD:

In the figure drawn above,

>>>extend ED upwards some more, up to point G ...above hypotenuse AC.

>>>extend FD to the left some more, up to point H ... beyond AC.

angle FCV = angle GDC = 40 degrees

angle EAD = angle HDA = 25 degrees

angle GDH = 90 degrees

so, angle CAD = 40 +90 +25 = 155 degrees. ----answer.