Question

27. In a AABC, B = 90, AB = 12 cm and BC = 5 cm.
Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C​

Answer 1

Given:-

• ∠B = 90°
• AB = 12cm
• BC = 5cm

Find:-

• cos A
• cosec A
• cos C
• cosec C

Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put( - 0.13,2.5){\large\bf 12cm}\put(2.8,.6){\large\bf 5cm}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\end{picture}$

Solution:-

In ∆ABC

$\underline{\boxed{\sf H^2 = P^2 + B^2}}\qquad\bigg\lgroup \sf Using\: Pythogoras\: Theorem\bigg\rgroup$

$\implies\sf AC^2 = AB^2 + BC^2$

$\sf where \small{\begin{cases} \sf AB = 12cm \\ \sf BC = 5cm \end{cases}}$

✯Substituting these values:-

$\dashrightarrow\sf AC^2 = AB^2 + BC^2$

$\dashrightarrow\sf AC^2 = 12^2 + 5^2$

$\dashrightarrow\sf AC^2 = 144 + 25$

$\dashrightarrow\sf AC^2 = 169$

$\dashrightarrow\sf AC= \sqrt{169}$

$\dashrightarrow\sf AC= 13cm$

Now, for T-ratios of ∠A.

Base = AB

Perpendicular = BC

$\sf (i) \cos A = \dfrac{AB}{AC}$

$\sf where \small{\begin{cases} \sf AB = 12cm \\ \sf AC = 13cm \end{cases}}$

♡Substituting these values:-

$\sf : \to\cos A = \dfrac{AB}{AC}$

$\sf : \to\cos A = \dfrac{12}{13}$

$\underline{\boxed{\sf \cos A = \dfrac{12}{13}}}$

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$\sf (ii) \cosec A = \dfrac{1}{ \sin A}$

$\sf\to \cosec A = \dfrac{1}{\dfrac{BC}{AC}}$

$\sf\to \cosec A =\dfrac{AC}{BC}$

$\sf where \small{\begin{cases} \sf AC=13cm \\ \sf BC = 5cm \end{cases}}$

☯Substituting these values:-

$\sf :\to \cosec A =\dfrac{AC}{BC}$

$\sf :\to \cosec A =\dfrac{13}{5}$

$\underline{\boxed{\sf \cosec A =\dfrac{13}{5}}}$

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Now, for T-ratio ∠A.

Base = BC

Perpendicular = AB

$\sf (iii) \cos C = \dfrac{BC}{AC}$

$\sf where \small{\begin{cases} \sf AC=13cm \\ \sf BC = 5cm \end{cases}}$

Substituting these values:-

$\sf \to \cos C = \dfrac{BC}{AC}$

$\underline{\boxed{\sf \cos C = \dfrac{5}{13}}}$

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$\sf (iv) \cosec C =\dfrac{1}{ \sin A}$

$\sf\to \cosec A = \dfrac{1}{\dfrac{AB}{AC}}$

$\sf\to \cosec A =\dfrac{AC}{AB}$

$\sf where \small{\begin{cases} \sf AB = 12cm \\ \sf AC = 13cm \end{cases}}$

Substituting these values:-

$\sf\to \cosec A =\dfrac{AC}{AB}$

$\sf\to \cosec A =\dfrac{13}{12}$

$\underline{\boxed{\sf \cosec A =\dfrac{13}{12}}}$

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