Question

In a right angled triangle abc if angle b =90, ac= 2 root 5 and ab-bc= 2 then, what is the value of cos square a- cos squarec

Answer 1

hope it will help you

Answer 2

Answer:

$\frac{3}{5}$

Step-by-step explanation:

In the question,

We have a right angled triangle ABC at B.

AC = 2√5

AB - BC = 2

We need to find the value of,

cos²A - cos²C

So,

We can say,

AB = x + 2

BC = x

In the triangle using Pythagoras Theorem,

AC² = AB² + BC²

So,

$(2\sqrt{5})^{2}=(x+2)^{2}+(x)^{2}\\20=2x^{2}+4+4x\\2x^{2}+4x=16\\x^{2}+2x-8=0\\(x+4)(x-2)=0\\x=-4,2$

So,

x = 2

Now,

AB = 4

and,

BC = 2

So,

$cos^{2}A=(\frac{4}{2\sqrt{5}})^{2}\\cos^{2}A=\frac{4}{5}$

also.

$cos^{2}C=(\frac{2}{2\sqrt{5}})^{2}\\cos^{2}C=\frac{1}{5}$

So,

$cos^{2}A-cos^{2}C=\frac{4}{5}-\frac{1}{5}\\cos^{2}A-cos^{2}C=\frac{3}{5}$

Therefore, the required value is,

$\frac{3}{5}$