Question

in traingle A=3 b= 5 and c=7 find the value of cosA cosB cosC ​

Answer 1

Answer:

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Answer 2

Question:-

• In triangle A=3 b= 5 and c=7 find the value of cosA cosB cosC

Given:-

1. A = 3
2. B = 5
3. C = 7

To Find:-

1. CosA
2. CosB
3. CosC

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Solution:-

$\boxed{ \sf \blue{</strong><strong>For,</strong><strong> </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>CosA </strong><strong>}}$

$\begin{gathered}\begin{gathered}:\red \implies\underline{ \boxed{\displaystyle \sf \bold{\: \cos(a) \: = \: \frac{ {b}^{2} \: + \: {c}^{2} \: - \: {a}^{2} }{2bc} }} }\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \:\cos(a) \: = \: \frac{ {5}^{2} \: + \: {7}^{2} \: - \: {3}^{2} }{2 \: \times \: 5 \: \times \: 7} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \:\cos(a) \: = \: \frac{ 25 \: + \: 49 \: - \: 9 }{2 \: \times \: 5 \: \times \: 7} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \underline\red{ \boxed{\displaystyle \sf \bold\green{\: \cos(a) \: = \: \frac{13}{14} }} }\\ \\\end{gathered}\end{gathered}$

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$\boxed{ \sf \blue{For, \: </strong><strong>Cos</strong><strong> </strong><strong>B</strong><strong>}}$

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: \cos(b) \: = \: \frac{ {c}^{2} \: + \: {a}^{2} \: - \: {b}^{2} }{2ac} }} }\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \:\cos(b) \: = \: \frac{ {7}^{2} \: + \: {3}^{2} \: - \: {5}^{2} }{2 \: \times \: 5 \: \times \: 7} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \:\cos(b) \: = \: \frac{ 49 \: + \: 9 \: - \: 25 }{2 \: \times \: 3 \: \times \: 7} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \underline\red{ \boxed{\displaystyle \sf \bold\green{\: \cos(b) \: = \: \frac{11}{14} }} }\\ \\\end{gathered}\end{gathered}$

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$\boxed{ \sf \blue{For, \: Cos C}}$

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: \cos(c) \: = \: \frac{ {a}^{2} \: + \: {b}^{2} \: - \: {c}^{2} }{2bc} }} }\\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \:\cos(c) \: = \: \frac{ {3}^{2} \: + \: {5}^{2} \: - \: {7}^{2} }{2 \: \times \: 3 \: \times \: 5} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \displaystyle \sf \:\cos(c) \: = \: \frac{ 9 \: + \: 25 \: - \: 49 }{2 \: \times \: 3 \: \times \: 7} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \underline\red{ \boxed{\displaystyle \sf \bold\green{\: \cos(b) \: = \: - \frac{1}{2} }} }\\ \\\end{gathered}\end{gathered}$

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Therefore:-

CosA = $\sf\blue{\frac{13}{14} }$

CosB = $\sf\blue{\frac{11}{14} }$

CosC = $\sf\blue{\frac{1}{2} }$

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