Question

In triangle ABC, right angled at B, AC=20cm and AB+BC=28cm. find sin A, cos A, sin A+ cos A.

Answer 1

Hi
I think what bithers you here is finding the lengths of each side of the triangle.
I assume you have basic knowledge of sin , cos and tan so that onve we finished finding the lengths of the sides..you can manage on your own :)

now since its a right angle triangle..we can apply pythagoras therom..in other words

(AB)^2 + (BC)^2 = (AC)^

now...AC = 20..therefor

(AB)^2 + (BC)^2 = (20)^2

we also know that

AB + BC = 28
then
BC = 28 - AB

therefor we can write

(AB)^2 + (28 - AB)^2 = (20)^2

lets simplify this one

(AB)^2 + (28)^2 + (AB)^2 -2(28AB) =(20)^2

2(AB)^2 -56AB +(28)^2 -(20)^2 = 0

we can use difference of 2 squares and write

(28)^2 - (20)^2 = (28 -20)(28 +20)
= 8(48)

now

2(AB)^2 -56AB +8(48) = 0

divide by 2

(AB)^2 -28AB +4(48)= 0

lets try to factorize this one ..

(AB)^2 -28AB +4(48)= 0

(AB)^2 -12AB - 16 AB +4(48)= 0

AB ( AB - 12) - 16 ( AB - 12) =0

(AB - 12)(AB - 16) =0

THEREFOR
AB = 12 or AB = 16

since the question does nit specify..which side is longer when compare AB & BC ..both 12 & 16 satisfy the side AB

when AB is 12..BC is 16
when AB is 16..BC is 12

we hust have to write 2 sets of sin cos and tan..when AB is equal to 12 and 16 repectively

sin A = opposite side / hypotenuse = BC / AC
cos A = adjecent side / hypotenuse = AB /AC
tan A = opposite side / adjecent side = BC /AB

hope this will.help you :)