Tolong ya yang bisa , harap bantuannya , nomor 13,14,15, dan 16.... Question from @amiramiwa

October 2018 1 18 Report

Tolong ya yang bisa , harap bantuannya , nomor 13,14,15, dan 16

hendrisyafa 13. $cos A= \frac{5^{2}+6^{2}- \sqrt{21}^{2} }{2.5.6}= \frac{25+36-21}{2.5.6}= \frac{40}{2.5.6}= \frac{2}{3}$

$Cos B = \frac{5^{2}+ \sqrt{21}^{2}-6^{2} }{2.5. \sqrt{21} }= \frac{25+21-36}{2.5. \sqrt{21} }= \frac{10}{10 \sqrt{21} }= \frac{1}{ \sqrt{21} }$

$Cos C = \frac{6^{2}+ \sqrt{21}^{2}-5^{2} }{2.6. \sqrt{21} } = \frac{36+21-25}{2.6. \sqrt{21} }= \frac{32}{2.6. \sqrt{21} }= \frac{8}{3 \sqrt{21} }$

Sehingga
Sin A = $\frac{ \sqrt{3^{2}-2^{2}} }{3}= \frac{1}{3} \sqrt{5}$
Sin B = $\frac{ \sqrt{ \sqrt{21}^{2}-1^{2} } }{ \sqrt{21} }= \frac{4}{21} \sqrt{105}$

Sin C = $\frac{ \sqrt{3 \sqrt{21}^{2}-8^{2} } }{3 \sqrt{21} }= \frac{189-64}{3 \sqrt{21} }= \frac{125}{3 \sqrt{21} }$

sudut sin terkecil = sudut A = 1/3 √5

14. $Cos A= \frac{b^{2}+c^{2}-a^{2}}{2.b.c}= \frac{7^{2}+8^{2}-9^{2}}{2.7.8}= \frac{49+64-81}{2.7.8}= \frac{32}{2.7.8}= \frac{2}{7}$

sehingga
Sin A = $\frac{ \sqrt{7^{2}-2^{2}} }{7}= \frac{ \sqrt{45} }{7}= \frac{3}{7} \sqrt{5}$

15. $cos\ \textless \ BAC = \frac{AB^{2}+AC^{2}-BC^{2}}{2.AB.BC}= \frac{7^{2}+6^{2}-5^{2}}{2.7.6}= \frac{49+36-25}{2.7.6}= \frac{60}{2.7.6}= \frac{5}{7}$

sehingga

Sin <BAC = $\frac{ \sqrt{7^{2}-5^{2}} }{7}= \frac{ \sqrt{24} }{7}= \frac{2}{7} \sqrt{6}$

16. Cos (A-B) = cos A cos B + sin A sin B
   = cos A cos B + 1/4 ...pers 1

jika A+B = π/3
maka cos (A+B) = cos (π/3)
   = 1/2
cos (A+B) = cos A cos B - sin A sin B
   1/2 = cos A cos B - 1/4

       cos A cos B = 1/2 + 1/4
   = 3/4 ......> subtitusikan ke pers 1
sehingga

cos (A-B) = 3/4 + 1/4
   = 1

1 votes Thanks 1

hendrisyafa done

amiramiwa terimakasih banyak atas jawabannya (:

hendrisyafa sm2 , expresinya " (: "

amiramiwa emm, maksudnya ' :) ' tadi hehe