Question

if the sides of triangle ABC are in the ratio 4:5:6 prove that one angle is twice of the other

Answer 1

We have in a triangle $\Delta ABC$,

Let $a=4x,b=5x,c=6x$ so that $a:b:c=4:5:6$.

[tex]\cos A=\frac{5^2+6^2-4^2}{2*5*6} =\frac{3}{4} \\ \cos B=\frac{6^2+4^2-5^2}{2*6*4} =\frac{9}{16} \\ \cos C=\frac{4^2+5^2-6^2}{2*4*5} =\frac{1}{8}[/tex].

We can see that

$\cos 2A=2 \cos ^2 A-1=2(\frac{3}{4} )^2-1=\frac{1}{8} =\cos C$.

That is

$C=2A$.

One angle is twice the other. Proof is complete.

Answer 2

Answer:

Step-by-step explanation:

by sine rule

sin A/a= sin B/b= sinC /c

⇒sinA/4=sinB/5=sinC/6

⇒sinA ratio sinB raitio sin C is 20:15:10

then angle A will be twice of angle C

hence proved