# A plank (AC) is leaning against a wall. The bottom of the plank(A) is 3 ft. from the wall- (AB = 3ft). Find AC and BC. There is a right angle at (angle ABC) and the angle of inclination of...

A plank (AC) is leaning against a wall. The bottom of the plank(A) is 3 ft. from the wall- (AB = 3ft). Find AC and BC. There is a right angle at (angle ABC) and the angle of inclination of A (ie angle CAB)is 49 degrees. Show work

Find AC (length of plank-hypoteneuse)

find BC height (off the ground where board touches the wall)

right angle (ABC) Angle CAB = 49 degrees

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Let's assess what we are given:

`AB=3` ie distance between the bottom of the plank & the wall

`/_CAB=49degrees` ie the angle formed between the ground and the plank

`AC= r` (unknown at this stage)

Draw a diagram to help you.

we know the trigonometric ratio for cosine cos 49=adj/hyp and we have the adjacent line (AB=3)

`therefore cos49=3/(AC)`

To find AC we must rearrange the equation that we have created:

`therefore cos 49 times AC = 3`

`therefore AC=3/(cos49)`

Now solve:

`therefore AC=3/0.656` (rounded off)

`therefore AC=4.57` ft

For BC we can use pythagorus or the same method as we just did using trig ratios. Thus:

`sin49=(BC)/(AC)`

`therefore sin 49 = (BC)/4.57`

`therefore4.57 times sin 49 = BC`

`therefore BC=3.45` ft (rounded off)

**Ans: AC=4.57 ft and BC = 3.45 ft**

If the angle of 49 degrees was the angle formed between the top of the wall and the ladder then your answer would be 3.98ft as you said and the calculation would have been

`sin /_ACB = 3/(AC)`

`therefore sin 49 = 3/(AC)`

`therefore sin 49 times AC = 3`

`therefore AC = 3/(sin 49)`

`therefore AC=3.98` ft and then of couse BC would change to 2.61ft

So it really depends where your angle of 49 degrees is. My understanding was that it existed where the ladder formed an angle with the ground (`/_CAB` ) but if it was the angle formed with the top of the wall (`/_ACB` ) then your answer of 3.98ft and 2.61 ft would be correct.