In a `DeltaABC`, if a = 18, b = 24, c = 30, find
(i) `sinA, sinB, sinC`
(ii) `cosA, cosB, cosC`

Explanation

Here, `a^(2)+b^(2)=(18)^(2)+(24)^(2)=(342+576)=900=(30)^(2)=c^(2).`
`:." "DeltaABC" is right-angled at C. thus, "/_C=90^(@).`
(i) By sine rule, we have
`a/("sin A")=b/("sin B")=c/("sin C")`
`rArr" "("sin A")/a=("sin B")/b=("sin C")/c="k (say)"`
`rArr" "("sin A")/18=("sin B")/24=("sin C")/30=k`
`:." "sinA=18k, sinB=24kandsinC=30k.`
`"But, "/_C=90^(@)rArrsinC=sin90^(@)rArr30k=1rArrk=1/30.`
`:." "sinA=18k=(18xx1/30)=3/5,B=24k=(24xx1/30)=4/5`
`"and sin C"==30k=(3-xx1/40)=1`.
Hence, `sinA=3/5,sinB=4/5andsinC=1`
(ii) Using cosine formulae, we get
`cosA=(b^(2)+a^(2)-b^(2))/(2ca)=((30)^(2)+(18)^(2)-(24)^(2))/(2xx24xx30)=(576+900-324)/(1440)=(1152)/(1440)=4/5`
`cosB=(c^(2)+a^(2)-b^(2))/(2ca)=((30)^(2)+(18)^(2)-(24)^(2))/(2xx30xx18)=((900+324-576))/(1080)=648/1080=3/5.`
`cosC=cos90^(@)=0`.
Hence, `cosA=4/5,cosB=3/5andcosC=0`