in triangle ABC AB=AC=15 and BC=18 then find cos B sin C
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I hope this is right and helpful
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Santhosh11111:
In my book answers are 3÷5
oh sorry I was thinking that it was a right triangle. use this formula b^2=c^2+a^2-2ac CosB and this will give you answer 3/5
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in ∆ ABC, AB=AC=15cm, BC =18cm.
AB=c , AC=b and BC=a
→ c=b=15 cm and a =18cm.
find ∠ABC or ∠B?
As per cosine formula of ∆s
cos∠B=(a²+c²−b²)/(2ac)=(18²+15²−15²)/(2...‡
=(18²)/(2×18×15) =3/5=0.6
AB=c , AC=b and BC=a
→ c=b=15 cm and a =18cm.
find ∠ABC or ∠B?
As per cosine formula of ∆s
cos∠B=(a²+c²−b²)/(2ac)=(18²+15²−15²)/(2...‡
=(18²)/(2×18×15) =3/5=0.6
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