Question

*ii)
If a :b:c=7:8:9, find cos A : cos B: cos C. ​

Answer 1

Given:

a:b:c=7:8:9

To find:

CosA:CosB:CosC

Explanation:

$\sf In \bigtriangleup ABC ,a:b:c=7:8:9$

$\sf let\: a=7k,b=8k,c=9k$

$\boxed{\sf CosA:CosB:CosC=a(b^2+c^2-a^2):b(a^2+c^2-b^2):c(a^2+b^2-c^2)}$

$\longrightarrow \: \sf \: a( {b}^{2} + {c}^{2} - {a}^{2} ) \\ \\ \longrightarrow \: \sf \: 7k( {(8k)}^{2} + {(9k)}^{2} - {(7k)}^{2} \\ \\ \longrightarrow \: \sf \: 7k(64 {k}^{2} + 81 {k}^{2} - 49 {k}^{2} ) \\ \\ \longrightarrow \: \sf \: 7k(96 {k}^{2} ) = 672 {k}^{3}$

$\longrightarrow \: \sf \: 8k( {(7k)}^{2} + {(9k)}^{2} - {(8k)}^{2} ) \\ \\ \longrightarrow \: \sf \: 8k(49 {k}^{2} + 81 {k}^{2} - 64 {k}^{2} ) \\ \\ \longrightarrow \: \sf \: 8k(66 {k}^{2} ) = 528 {k}^{3}$

$\longrightarrow \: \sf \: 9k( {(7k)}^{2} + {(8k)}^{2} - {(9k)}^{2} ) \\ \\ \longrightarrow \: \sf \: 9k(49 {k}^{2} + 64 {k}^{2} - 81 {k}^{2} ) \\ \\ \longrightarrow \: \sf \: 9k(32 {k}^{2} ) = 288 {k}^{3}$

$\sf Ratio=672k^3:528k^3:288k^3$

$\longrightarrow \: \sf 672:528:288$

$\longrightarrow \: \sf 336:264:144$

$\longrightarrow \: \sf 168:132:72$

$\longrightarrow \: \sf 84:66:36$

$\longrightarrow \: \sf 42:33:18$

$\longrightarrow \: \sf 14:11:6$

$\boxed{\sf CosA:CosB:CosC=14:11:16}$

Answer 2

Answer:

I think this might help u