Question

In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) Sin A, cos A
(ii) sin C, cos C
Sol. In right ΔABC, we have: p = 24 cm, b = 7 cm
(by using pythgoras theorem)
(class 10,cbse)
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Answer 1

Appropriate Question:

• In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
• (i) Sin A, cos A
• (ii) sin C, cos C

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$\star$ DIAGRAM:

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.1,2){\sf{\large{24 cm}}}\put(9,0.7){\sf{\large{7 cm}}}\put(9.4,1.9){\sf{\large{25 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

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In the given right ∆ABC,

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$\underline{\bigstar\:\boldsymbol{Using\: Pythagoras \ theorem\::}}$

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$\star\;\boxed{\sf{\pink{(Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2}}}$

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$:\implies\sf AC^2 = AB^2 + BC^2 \\\\\\:\implies\sf AC^2 = (24)^2 + (7)^2 \\\\\\:\implies\sf AC^2 = 576 + 49\\\\\\:\implies\sf AC = \sqrt{625}\\\\\\:\implies {\underline{\boxed{\frak{\purple{AC = 25\: cm}}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ value \ of \ AC \ is \ \bf{25 \ cm}.}}}$

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I) Sin A, Cos A

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$:\implies\sf Sin \ A = \dfrac{Base}{Hypotenuse}\\\\\\:\implies\sf Sin \ A = \dfrac{BC}{AC}\\\\\\:\implies{\underline{\boxed{\frak{\pink{Sin \ A = \dfrac{7}{25}}}}}}\:\bigstar$

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$:\implies\sf Cos \ A = \dfrac{Perpendicular}{Hypotenuse}\\\\\\:\implies\sf Cos \ A = \dfrac{AB}{AC}\\\\\\:\implies{\underline{\boxed{\frak{\pink{Cos \ A = \dfrac{24}{25}}}}}}\:\bigstar$

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II) Sin C, Cos C⠀⠀

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$:\implies\sf Sin \ C = \dfrac{Perpendicular}{Hypotenuse}\\\\\\:\implies\sf Sin \ C = \dfrac{AB}{AC}\\\\\\:\implies{\underline{\boxed{\frak{\purple{Sin \ C = \dfrac{24}{25}}}}}}\:\bigstar$

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$:\implies\sf Cos \ C = \dfrac{Base}{Hypotenuse}\\\\\\:\implies\sf Cos \ C = \dfrac{BC}{AC}\\\\\\:\implies{\underline{\boxed{\frak{\purple{Cos \ C = \dfrac{7}{25}}}}}}\:\bigstar$

Answer 2

✅ See the attachment diagram.

$\\ \Large{\bf{\green{\underline{GiVeN\::}}}}$

• ABC is a right angle triangle

• ∠B is right angled, i.e. 90°.

• AB (Height) = 24 cm

• BC (Base) = 7 cm

$\\ \Large{\bf{\pink{\underline{To\:FiNd\::}}}}$

1. Sin A , Cos A
2. Sin C , Cos C

$\\ \Large{\bf{\purple{\underline{CaLcUlAtIoN\::}}}}$

$\bf\orange{According\:to\:Pythagoras\:theorem,}$

$\red\bigstar\:\:{\underline{\blue{\boxed{\bf{\green{(Hypotenuse)^2\:=\:(Height)^2\:+\:(Base)^2\:}}}}}}$

$:\implies\:\:\bf{(Hypotenuse)^2\:=\:(24)^2\:+\:7^2\:}$

$:\implies\:\:\bf{(Hypotenuse)^2\:=\:576\:+\:49\:}$

$:\implies\:\:\bf{(Hypotenuse)^2\:=\:625\:}$

$:\implies\:\:\bf{Hypotenuse\:=\:\sqrt{625}\:}$

$:\implies\:\:\bf\blue{Hypotenuse\:(AC)\:=\:25\:cm}$

$\bf\pink{We\:know\:that,}$

$\orange\checkmark\:\:{\underline{\boxed{\bf{\purple{Sin\:\theta\:=\:\dfrac{Perpendicular}{Hypotenuse}}}}}}$

• Perpendicular is also the opposite side.

$\green\checkmark\:\:{\underline{\boxed{\bf{\blue{Cos\:\theta\:=\:\dfrac{Base}{Hypotenuse}}}}}}$

• Base is also the adjacent side.

❶ Sin A , Cos A

$:\implies\:\:\bf{Sin\:A\:=\:\dfrac{BC}{AC}\:}$

$:\implies\:\:\bf\red{Sin\:A\:=\:\dfrac{7}{25}\:}$

$\Large\bf\purple{And,}$

$:\implies\:\:\bf{Cos\:A\:=\:\dfrac{AB}{AC}\:}$

$:\implies\:\:\bf\pink{Cos\:A\:=\:\dfrac{24}{25}\:}$

❷ Sin C , Cos C

$:\implies\:\:\bf{Sin\:C\:=\:\dfrac{AB}{AC}\:}$

$:\implies\:\:\bf\orange{Sin\:C\:=\:\dfrac{24}{25}\:}$

$\Large\bf\blue{And,}$

$:\implies\:\:\bf{Cos\:C\:=\:\dfrac{BC}{AC}\:}$

$:\implies\:\:\bf\green{Cos\:C\:=\:\dfrac{7}{25}\:}$