Question

If angle A and angle B are acute angle such that cosA= cosB, then show that angleA= angle B​

Answer 1

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Answer 2

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$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{A}}\put(7.7,1){\large{C}}\put(10.6,1){\large{B}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

$\huge{\mathfrak{\green{\red{\bf\:S}\pink{o}\purple{l}\green{u}\orange{t}\blue{i}\blue{o}\red{n:-}}}}$

$\rm\:Let\:ACB\:is\:a\:right\:triangle$

$\rm\:{\angle{C}}=90\degree$

Now,

$\rm\:cosA=\dfrac{b}{h}=\dfrac{AC}{AB}$

$\rm\:cosB=\dfrac{b}{h}=\dfrac{BC}{AB}$

$\rm\therefore\:cosA=cosB\:\:(given)$

$\rm\implies\dfrac{AC}{\cancel{AB}}=\dfrac{BC}{\cancel{AB}}$

$\rm\implies\:AC=BC$

$\rm\implies\:{\angle{B}}={\angle{A}}$

$\rm\implies\:{\angle{A}}={\angle{B}}$

$\rm\:Hence,Proved$

Some formula of Trigonometry:-

• sinθ = p/h

• cosθ = b/h

• tanθ=p/b

• cotθ=b/p

• secθ=h/b

• cosecθ=h/p