# Thread: help me solve this considered easy maths question

1. ## help me solve this considered easy maths question

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

2. Originally Posted by helloying
the AC is not a straight line by the way. Just afraid my drawings are not clear
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:

$\sin 40^o = \frac E8$

$\cos 25^o = \frac F9$

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.

3. oh i see thanks for ur help. u are so smart.

4. Originally Posted by helloying
oh i see thanks for ur help. u are so smart.
please note that i have edited my post, i forgot about the second part of the question

5. Originally Posted by helloying
the AC is not a straight line by the way. Just afraid my drawings are not clear
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.

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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.

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