In a right angle triangle ABC with right angle at B, in which a=24 units, b=25 units and 11012BAC = 0. Then, find cos 6 and tan e. 0
Answers
Answer:
\huge \bf {Answer:-}Answer:−
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\sf \red {\underline{Given:-}}
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:-}}
Tofind:−
(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}}→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}}→(AC)
2
=(BC)
2
+(AB)
2
\to \tt {(25)^{2}=(24)^{2}+(AB)^{2}}→(25)
2
=(24)
2
+(AB)
2
\to \tt {625=576+AB^{2}}→625=576+AB
2
\to \tt {AB^{2}=625-576}→AB
2
=625−576
\to \tt {AB^{2}=49}→AB
2
=49
\to \tt {AB=\sqrt{49}}→AB=
49
\to \tt {\fbox{AB=7}}→
AB=7
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\sf \pink {\underline{Now,}}
Now,
(i)Cos θ=?
★We know,
\to \tt {Cos\:θ=\frac{Base}{Hypotenuse}}→Cosθ=
Hypotenuse
Base
★From the figure,we can observe
Hypotenuse=AC
Base=AB
\to \tt {Cos\:θ=\frac{AB}{AC}}→Cosθ=
AC
AB
\large \implies \green {Cos\:θ=\frac{7}{25}}⟹Cosθ=
25
7
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(ii)Tan θ=?
\sf \purple {\underline{We\: know,}}
Weknow,
\to \tt {Tan\:θ=\frac{perpendicular}{Base}}→Tanθ=
Base
perpendicular
perpendicular=BC
Base=AB
\to \tt {Tan\:θ=\frac{BC}{AB}}→Tanθ=
AB
BC
\large \implies \green{Tan\:θ=\frac{24}{7}}⟹Tanθ=
7
24
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\sf \orange {\underline{Thence,}}
Thence,
★The values of Cos θ and Tan θ are {\frac{7}{25}and{\frac{24}{7}}}
25
7
and
7
24
respectively.
Answer:
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