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This is one of our many quizzes on Law of cosines. Click on this to access more such competitive quizzes on Trigonometry.

Law of cosines Quiz 2

mathematics

 What is $\sqrt{7}$ times the reciprocal of the value of $\cos (\beta )$ if $a=3,b=2,c=4$.

 Apply law of cosines and find $\cos (\beta )$. Then find the reciprocal and multiply the value.

  Using law of cosines, $ \\ {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cdot \cos \left( \beta  \right) \\ \Rightarrow \cos \left( \beta  \right)=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} \\ \cos \left( \beta  \right)=\dfrac{{{3}^{2}}+{{4}^{2}}-{{2}^{2}}}{2\left( 3 \right)\left( 4 \right)} \\ \text{           }=\dfrac{9+16-4}{24}=\dfrac{7}{8} \\ \therefore \cos \left( \beta  \right)=\dfrac{7}{8} \\ $ The reciprocal of $\cos (\beta )$is $\dfrac{8}{7}$. And $\sqrt{7}$ times $\dfrac{8}{7}$ is $\dfrac{8}{\sqrt{7}}$

 The sides of a triangle are given by  $b=12$ and $c=2$ .If the side $c$ is squared then what will be side $a$ if $\beta =90{}^\circ $.

 Apply law of cosines directly, be careful about which two sides have the angle $\beta $. Don’t forget to square ‘c’.

  Take $c=4$, as c is squared. Apply law of cosine formula, $ \\ 12=\sqrt{{{a}^{2}}+{{4}^{2}}-2a(4)\cos (90{}^\circ )} \\ \Rightarrow 144={{a}^{2}}+16-0 \\ \Rightarrow {{a}^{2}}=128 \\ \Rightarrow a=8\sqrt{2} $

 What is the ratio of the side $a$ when$b=7$,$c=8$ and angle $\alpha =\pi $ and $b=3$,$c=4$ and angle given as $\alpha =\dfrac{3\pi }{2}$

 Apply law of cosines directly. Find the side $a$ and the determine the ratio. 

  When $b=7$,$c=8$ and angle $\alpha =\pi $, Then $ \\ {{a}^{2}}=\sqrt{{{b}^{2}}+{{c}^{2}}-2bc\cos (\alpha )} \\ {{a}^{2}}=\sqrt{{{7}^{2}}+{{8}^{2}}-2\left( 7 \right)\left( 8 \right)\cos \left( \pi  \right)} \\ \Rightarrow {{a}^{2}}=\sqrt{49+64-112\left( -1 \right)} \\ \Rightarrow {{a}^{2}}=225 \\ \Rightarrow a=15 \\ $ When  $b=3$,$c=4$ and angle given as $\alpha =\dfrac{3\pi }{2}$. $ \\ {{a}^{2}}=\sqrt{{{b}^{2}}+{{c}^{2}}-2bc\cos (\alpha )} \\ {{a}^{2}}=\sqrt{{{3}^{2}}+{{4}^{2}}-2\left( 3 \right)\left( 4 \right)\cos \left( \dfrac{3\pi }{2} \right)} \\ \Rightarrow {{a}^{2}}=\sqrt{9+16-24\left( 0 \right)} \\ \Rightarrow {{a}^{2}}=25 \\ \Rightarrow a=5 \\ \therefore $ The ratio is $\dfrac{15}{5}=3$

 If D is the midpoint of AB. Then find $\dfrac{3C{{D}^{2}}}{4B{{C}^{2}}} \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/dabc5417-aefb-435d-b283-9e9932d9a4257487344591732620582.png"/>

 Use Pythagoras theorem to find the unknown term, then use the values in the given expression.

  Applying Pythagoras theorem for $\Delta ADC$, $ \\ C{{D}^{2}}=16+16=32 \\ $ Applying Pythagoras theorem for $\Delta ABC$, $ \\ B{{C}^{2}}=16+{{(4+4)}^{2}} \\ \Rightarrow B{{C}^{2}}=16+64=80 \\ $ Substituting values, $ \\ \dfrac{3C{{D}^{2}}}{4B{{C}^{2}}}=\dfrac{3}{4}\times \dfrac{32}{80}=\dfrac{3}{10}$

 Find  $A{{D}^{2}}+B{{C}^{2}}$ for the given figure if $\angle ABD=60{}^\circ $ and $\angle BDC=30{}^\circ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5c981e46-4e8f-449c-adce-09ac12deb5dc1482821032170987622.png"/>

 Apply law of cosines directly to different triangles, be careful about which two sides have the angle the corresponding angles given. Add corresponding values to obtain the answer.

  Consider $\Delta ABD$ and apply law of cosines formula, we get $ \\ A{{D}^{2}}={{4}^{2}}+{{5}^{2}}-2(4)(5)\cos (60{}^\circ ) \\ \Rightarrow A{{D}^{2}}=41-40\times \dfrac{1}{2} \\ \therefore A{{D}^{2}}=21 \\ $ Consider $\Delta BDC$ and apply law of cosines formula, we get $ \\  B{{C}^{2}}={{2}^{2}}+{{5}^{2}}-2(2)(5)\cos (30{}^\circ ) \\ \Rightarrow B{{C}^{2}}=29-20\times \dfrac{\sqrt{3}}{2} \\ \therefore B{{C}^{2}}=29-10\sqrt{3} \\ $ So, $ \\ A{{D}^{2}}+B{{C}^{2}}=21+29-10\sqrt{3} \\ \Rightarrow A{{D}^{2}}+B{{C}^{2}}=10\left( 5-\sqrt{3} \right)$