Question

$\Delta$ABC, AD $\perp$ CB, AC $=29$ units, DC $=21$ units and AB $=40$ units. Find the value of ${\cos }^{2}B-{\sin }^{2}B$

Answer 1

R.E.F image

Given $AD\perp BC$ AC=29 unit

CD=21 unit AB=40unit

for, given figure

we have In $△ACD.$

$(29{)}^2=(21{)}^2+(AD{)}^2$ [Pythagoras theorem]

$(AD{)}^2=841-441=400$

[AD=20] In $△ADB.$

we know $(BD{)}^2=(AB{)}^2-(AD{)}^2=1600-400$

$(BD{)}^2$ =1200

BD=20$\sqrt{3}unit$

now cos B =$\frac{BD}{AB}$ =$\frac{20\sqrt{3}}{40}$=$\frac{\sqrt{3}}{2}$ sin B

then $co{s}^{2}B-si{n}^{2}B=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$