Question

In a right angled triangle ABC with right angle at B, in which a=24 units, b= 25 units and ⟨BAC=0. Then, find cos 0 and tan 0​

Answer 1

$\huge \bf {Answer:-}$

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$\sf \red {\underline{Given:-}}$

★ABC is a right angled triangle which is right angled at B

★a=24 units, b= 25 units and <BAC=θ

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$\sf \blue {\underline{To\: find:-}}$

(i)Cos θ=?

(ii)Tan θ=?

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★From the given info,we can observe , that two sides of the right-angled triangle are already given,we can find out the third side

★According to Pythagoras theorem,

$\to \tt {Hypotenuse^{2}=Side^{2}+Side^{2}}$

★Where, Hypotenuse is AC=25 units

★and one side BC =24 units

★other side AB=?

$\to \tt {(AC)^{2}=(BC)^{2}+(AB)^{2}}$

$\to \tt {(25)^{2}=(24)^{2}+(AB)^{2}}$

$\to \tt {625=576+AB^{2}}$

$\to \tt {AB^{2}=625-576}$

$\to \tt {AB^{2}=49}$

$\to \tt {AB=\sqrt{49}}$

$\to \tt {\fbox{AB=7}}$

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$\sf \pink {\underline{Now,}}$

(i)Cos θ=?

★We know,

$\to \tt {Cos\:θ=\frac{Base}{Hypotenuse}}$

★From the figure,we can observe

Hypotenuse=AC

Base=AB

$\to \tt {Cos\:θ=\frac{AB}{AC}}$

$\large \implies \green {Cos\:θ=\frac{7}{25}}$

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(ii)Tan θ=?

$\sf \purple {\underline{We\: know,}}$

$\to \tt {Tan\:θ=\frac{perpendicular}{Base}}$

perpendicular=BC

Base=AB

$\to \tt {Tan\:θ=\frac{BC}{AB}}$

$\large \implies \green{Tan\:θ=\frac{24}{7}}$

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$\sf \orange {\underline{Thence,}}$

★The values of Cos θ and Tan θ are ${\frac{7}{25}and{\frac{24}{7}}}$ respectively.