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MATHS 3 Marks Question

Sine And Cosine Formulae And Their Applications · MP Board (MPBSE) STD 11 Science (Madhya Pradesh - English Medium). 13 questions.

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Question 13 Marks
$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{c}\sec\text{C + a}\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\frac{\text{a}\sec\text{A + b}\sec\text{B}}{\tan\text{A}+\tan\text{B}}.$
Answer
$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{c}\sec\text{C + a}\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\frac{\text{a}\sec\text{A + b}\sec\text{B}}{\tan\text{A}+\tan\text{B}}$
$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}$
$=\frac{\text{k}\sin\text{B}\sec\text{B + k}\sin\text{C}\sec\text{C}}{\tan\text{B}+\tan\text{C}}$
$=\frac{\text{k}\sin\text{B}\frac{1}{\cos\text{B}}+\text{k}\sin\text{C}\frac{1}{\cos\text{C}}}{\tan\text{B}+\tan\text{C}}$
$=\frac{\text{k}\tan\text{B + k}\tan\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{k}(\tan\text{B}+\tan\text{C})}{\tan\text{B}+\tan\text{C}}=\text{k}$
Similarly, $\frac{\text{c}\sec\text{C}+\text{a}+\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\text{k}$
Similarly, $\frac{\text{a}\sec\text{A}+\text{b}+\sec\text{B}}{\tan\text{A}+\tan\text{B}}=\text{k}$
Hence, it is proved that
$\frac{\text{b}\sec\text{B + c}\sec\text{C}}{\tan\text{B}+\tan\text{C}}=\frac{\text{c}\sec\text{C + a}\sec\text{A}}{\tan\text{C}+\tan\text{A}}=\frac{\text{a}\sec\text{A + b}\sec\text{B}}{\tan\text{A}+\tan\text{B}}=\text{k}$
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Question 23 Marks
$\frac{\text{a}^2-\text{c}^2}{\text{b}^2}=\frac{\sin(\text{A}-\text{C})}{\sin(\text{A}+\text{C})}$
Answer
$\frac{\text{a}^2-\text{c}^2}{\text{b}^2}=\frac{\sin(\text{A}-\text{C})}{\sin(\text{A}+\text{C})}$
let $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$
$\text{LHS}=\frac{\text{a}^2-\text{c}^2}{\text{b}^2}$
$=\frac{\text{k}^2\sin^2\text{A}-\text{k}^2\sin^2\text{C}}{\text{k}^2\sin^2\text{B}}$
$=\frac{\text{k}^2(\sin^2\text{A}-\sin^2\text{C})}{\text{K}^2\sin^2\text{B}}$
$=\frac{(\sin^2\text{A}-\sin^2\text{C})}{\sin^2(\pi-(\text{A + C})}$
$=\frac{\sin(\text{A + C})\sin(\text{A} - \text{C})}{\sin^2(\text{A + C})}$
$=\frac{\sin(\text{A} - \text{C})}{\sin\text{(A + C)}}=\text{RHS}$
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Question 33 Marks
$\text{b}\sin\text{B}-\text{c}\sin\text{C = a}\sin(\text{B}-\text{C})$
Answer
$\text{b}\sin\text{B}-\text{c}\sin\text{C = a}\sin(\text{B}-\text{C})$
$\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\text{RHS}=\text{a}\sin(\text{B}-\text{C})$
$=\text{a}\sin\text{B}.\cos\text{C}-\text{a}\sin \text{C}.\cos\text{B}$
$=\text{a(bk).}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{a(ck)}.\Big(\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}\Big)$
$=\text{k}.\frac{(\text{a}^2+\text{b}^2-\text{c}^2)}{2}-\text{k}\frac{(\text{a}^2+\text{c}^2-\text{b}^2)}{2}$
$=2\text{k}.\frac{(\text{b}^2-\text{c}^2)}{2}$
$=\text{b.(kb)}-\text{c(kc)}$
$=\text{b}(\sin\text{B})-\text{c}(\sin\text{C})=\text{LHS}$
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Question 43 Marks
$\text{b}\cos\text{B + c}\cos\text{C}=\text{a}\cos(\text{B}-\text{C})$
Answer
$\text{b}\cos\text{B + c}\cos\text{C}=\text{a}\cos(\text{B}-\text{C})$
Let $\text{a = k}\sin\text{A,b = k}\sin\text{B,c = k}\sin\text{C}$
$\text{LHS}=\text{b}\cos\text{B + c}\cos\text{C}$
$=\text{k}\sin\text{B}\cos\text{B}+\text{k}\sin\text{C}\cos\text{C}$
$=\frac{\text{k}}{2}(2\sin\text{B}\cos\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(\sin2\text{B}+\sin2\text{C})$
$=\frac{\text{k}}{2}2\sin(\text{B + C})\cos(\text{B}-\text{C})$
$=\text{k}\sin(\pi-\text{A})\cos(\text{B}-\text{C})$
$=\text{k}\sin\text{A}\cos(\text{B}-\text{C})$
$=\text{a}\cos(\text{B}-\text{C})=\text{RHS}$
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Question 53 Marks
In a $\triangle\text{ABC},$ if a = 18, b = 24, c = 30, find $\cos\text{A},\cos\text{B}$ and $\cos\text{C}.$
Answer
In any $\triangle\text{ABC},$ we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{ab}}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
we have,
a = 18, b = 24, c = 30
Therefore,
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}=\frac{1152}{1440}=\frac{4}{5}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}=\frac{648}{1080}=\frac{3}{5}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=\frac{0}{864}=0$
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Question 63 Marks
$\frac{\text{c}-\text{b}\cos\text{A}}{\text{b}-\text{c}\cos\text{A}}=\frac{\cos\text{B}}{\cos\text{C}}$
Answer
$\frac{\text{c}-\text{b}\cos\text{A}}{\text{b}-\text{c}\cos\text{A}}=\frac{\cos\text{B}}{\cos\text{C}}$
$\text{LHS}=\frac{\text{c}-\text{b}\cos\text{A}}{\text{b}-\text{c}\cos\text{A}}$
$=\frac{\text{k}\sin\text{C}-\text{k}\sin\text{B}\cos\text{A}}{\text{k}\sin\text{B}-\text{k}\sin\text{C}\cos\text{A}}$
$=\frac{\sin(\pi-(\text{A + B}))-\sin\text{B}\cos\text{A}}{\sin(\pi(\text{A + C}))-\sin\text{C}\cos\text{A}}$
$=\frac{\sin(\text{A + B})-\sin\text{B}\cos\text{A}}{\sin(\text{A + C})-\sin\text{C}\cos\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}-\sin\text{B}\cos\text{A}}{\sin\text{A}\cos\text{C}+\cos\text{A}\sin\text{C}-\sin\text{C}\cos\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}}{\sin\text{A}\cos\text{C}}$
$=\frac{\cos\text{B}}{\cos\text{C}}=\text{RHS}$
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Question 73 Marks
$\text{a}^2(\cos^2\text{B}-\cos^2\text{C})+\text{b}^2(\cos^2\text{C}-\cos^2\text{A})\\+\text{c}^2(\cos^2\text{A}-\cos^2\text{B})=0$
Answer
$\text{a}^2(\cos^2\text{B}-\cos^2\text{C})+\text{b}^2(\cos^2\text{C}-\cos^2\text{A})\\+\text{c}^2(\cos^2\text{A}-\cos^2\text{B})=0$
$\text{LHS}=\text{a}^2(1-\sin^2\text{B}-1+\sin^2\text{C})\\\ \ \ \ \ \ \ \ \ \ \ +\text{b}^2(1-\sin^2\text{C}-1+\sin^2\text{A})\\ \ \ \ \ \ \ \ \ \ \ \ +\text{c}^2(1-\sin^2\text{A}-1+\sin^2\text{B})$
$=\text{a}^2(\sin^2\text{C}-\sin^2\text{B})+\text{b}^2(\sin^2\text{A}-\sin^2\text{C})\\+\text{c}^2(\sin^2\text{B}-\sin^2\text{A})$
$=\text{a}^2(\text{k}^2\text{c}^2-\text{k}^2\text{b}^2)+\text{b}^2(\text{k}^2\text{a}^2-\text{k}^2\text{c}^2)+\text{c}^2(\text{k}^2\text{b}^2-\text{k}^2\text{a}^2)$
$=\text{k}^2(\text{a}^2\text{c}^2-\text{a}^2\text{b}^2+\text{b}^2\text{a}^2-\text{b}^2\text{c}^2+\text{b}^2\text{c}^2-\text{a}^2\text{c}^2)$
$=\text{k}^2\times0 = 0=\text{RHS}$
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Question 83 Marks
$\frac{\text{a + b}}{\text{c}}=\frac{\cos\big(\frac{\text{A}-\text{B}}{2}\big)}{\sin\frac{\text{C}}{2}}$
Answer
$\text{LHS}=\frac{\text{a + b}}{\text{c}}$
Let $\text{a = k}\sin\text{A, b = k}\sin\text{B, c = k}\sin\text{C}$
$=\frac{\text{k}\sin\text{A}+\text{k}\sin\text{B}}{\text{k}\sin\text{C}}$
$=\frac{\text{}\sin\text{A}+\text{}\sin\text{B}}{\text{}\sin\text{C}}$
$=\frac{2\sin\frac{\text{A + B}}{2}.\cos\frac{\text{A}-\text{B}}{2}}{2\sin\frac{\text{C}}{2}.\cos\frac{\text{C}}{2}}$
$=\frac{\sin\big(\frac{\pi-\text{C}}{2}\big).\cos\frac{\text{A}-\text{B}}{2}}{\sin\frac{\text{C}}{2}.\cos\frac{\text{C}}{2}}$
$=\frac{\cos\frac{\text{A}-\text{B}}{2}}{\sin\frac{\text{C}}{2}}=\text{RHS}$
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Question 93 Marks
$\text{b}(\text{c}\cos\text{A}-\text{a}\cos\text{C})=\text{c}^2-\text{a}^2$
Answer
$\text{b}(\text{c}\cos\text{A}-\text{a}\cos\text{C})=\text{c}^2-\text{a}^2$$\text{RHS}=\text{c}^2-\text{a}^2$
$=\text{k}^2\sin^2\text{C}-\text{k}^2\sin^2\text{A}$
$=\text{k}^2(\sin^2\text{C}-\sin^2\text{A})$
$=\text{k}^2\sin(\text{C + A}).\sin(\text{C}-\text{A})$
$=\text{k}^2\sin(\pi-\text{B}).\sin(\text{C}-\text{A})$
$=\text{k}^2\sin\text{B}.\sin(\text{C}-\text{A})$
$=\text{k}\sin\text{B}.\text{k}\sin(\text{C}-\text{A})$
$=\text{bk}\sin(\text{C}-\text{A})$
$=\text{bk}(\sin\text{C}.\cos\text{A}-\sin\text{A}.\cos\text{C})$
$=\text{b}(\text{k}\sin\text{C}.\cos\text{A}-\text{k}\sin\text{A}.\cos\text{C})$
$=\text{b}(\text{c}\cos\text{A}-\text{a}\cos\text{C})=\text{LHS}$
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Question 103 Marks
$(\text{c}^2-\text{a}^2+\text{b}^2)\tan\text{A}=(\text{a}^2-\text{b}^2+\text{c}^2)\tan\text{B}=(\text{b}^2-\text{c}^2+\text{a}^2)\tan\text{C}$
Answer
For any $\triangle\text{ABC},$ we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
therefore,
$(\text{c}^2+\text{b}^2-\text{a}^2)\tan\text{A}=(\text{c}^2+\text{b}^2-\text{a}^2)\frac{\sin\text{A}}{\cos\text{A}}$
$=(\text{c}^2+\text{b}^2-\text{a}^2)\frac{\text{ka}}{\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}}$
$=2\text{kabc}$
Also,
$(\text{a}^2+\text{c}^2-\text{b}^2)\tan\text{B}=(\text{a}^2+\text{c}^2-\text{b})^2\frac{\sin\text{B}}{\cos\text{B}}$
$=(\text{a}^2+\text{c}^2-\text{b}^2)\frac{\text{kb}}{\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}}$
$=2\text{kabc}$
Now,
$(\text{a}^2+\text{b}^2-\text{c}^2)\tan\text{C}=(\text{a}^2+\text{b}^2-\text{c}^2)\frac{\sin\text{C}}{\cos\text{C}}$
$=(\text{a}^2+\text{b}^2-\text{c}^2)\frac{\text{kc}}{\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}}$
$=2\text{kabc}$
Hence proved.
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Question 113 Marks
$2(\text{bc}\cos\text{A}+\text{ca}\cos\text{B}+\text{ab}\cos\text{C})=\text{a}^2+\text{b}^2+\text{c}^2$
Answer
$2(\text{bc}\cos\text{A}+\text{ca}\cos\text{B}+\text{ab}\cos\text{C})=\text{a}^2+\text{b}^2+\text{c}^2$$\text{LHS}=2\text{bc}\cos\text{A}+2\text{ca}\cos\text{B}+2\text{ab}\cos\text{C}$
$=2\text{bc}\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}+2\text{ca}\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ca}}+2\text{ab}\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
$=\text{b}^2+\text{c}^2-\text{a}^2+\text{a}^2+\text{c}^2-\text{b}^2+\text{a}^2+\text{b}^2-\text{c}^2$
$=\text{a}^2+\text{b}^2+\text{c}^2=\text{RHS}$
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Question 123 Marks
$\text{c}(\text{a}\cos\text{B}-\text{b}\cos\text{A})=\text{a}^2-\text{b}^2$
Answer
$\text{c}(\text{a}\cos\text{B}-\text{b}\cos\text{A})$
$=\text{ac}.\cos\text{B}-\text{bc}\cos\text{A}$
$=\text{ac}.\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}-\text{bc}.\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2}-\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2}$
$=\frac{\text{a}^2+\text{c}^2-\text{b}^2-\text{b}^2-\text{c}^2+\text{a}^2}{2}$
$=\frac{2\text{a}^2-2\text{b}^2}{2}=(\text{a}^2-\text{b}^2)$
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Question 133 Marks
$\text{a}^2\sin(\text{B}-\text{C})=(\text{b}^2-\text{c}^2)\sin\text{A}$
Answer
$\text{a}^2\sin(\text{B}-\text{C})=(\text{b}^2-\text{c}^2)\sin\text{A}$
$\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\text{LHS}=\text{a}^2\sin(\text{B}-\text{C})$
$=\text{a}^2(\sin\text{B}.\cos\text{C}-\sin\text{C}.\cos\text{B})$
$=\text{a}^2\text{kb.}\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}-\text{a}^2\text{ck.}\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$ [Using cos rule and sine rule]
$=\text{a}^2\text{k}.\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{a}}-\text{a}^2\text{k}.\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{a}}$
$=\text{a}^2\text{k}.\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2-\text{a}^2-\text{c}^2+\text{b}^2}{2\text{a}}\Big)$
$=\text{a}^2\text{k}.\Big(\frac{2\text{b}^2-2\text{c}^2}{2\text{a}}\Big)$
$=\text{ak.(b}^2-\text{c}^2)$
$=\sin\text{A(b}^2-\text{c}^2)=\text{RHS}$
Hence Proved
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