A ship leaves harbour on a course N 72° E and after traveling for 50 metres, changes course to 108°. After a further 106 metres, find
(a) the distance of the ship from the harbour,
(b) its bearing from the harbour.
إظهار الرسائل ذات التسميات law of cosine. إظهار كافة الرسائل
إظهار الرسائل ذات التسميات law of cosine. إظهار كافة الرسائل
الأربعاء، 28 نوفمبر 2018
الثلاثاء، 27 نوفمبر 2018
Exercise (11.3)-Problem 17
Two runners start from the same point at 12:00 noon, one of them heading north at 6 mph and the other heading 68° east of north at 8 mph. What is the distance between them at 3:00 that afternoon?
الخميس، 22 نوفمبر 2018
Law of Cosines and to Find Extremum by Completing Square
In $ \displaystyle \Delta ABC$, $ \displaystyle AB=(5-x)$ cm, $ \displaystyle BC=(4+x)$ cm, $ \displaystyle \angle AsBC=120{}^\circ $ and $ \displaystyle AC=y$ cm.
(a) Show that $ \displaystyle {{y}^{2}}={{x}^{2}}-x+61$.
(b) Find the minimum value of $ \displaystyle {{y}^{2}}$, and give the value of $ \displaystyle x$ for which this occurs.
Solution
$ \displaystyle AB=(5-x)$ cm,
$ \displaystyle BC=(4+x)$ cm,
$ \displaystyle \angle ABC=120{}^\circ $
$ \displaystyle AC=y$ cm.
(a) By the law of cosines,
$ \displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2\cdot AB\cdot AC\cos (\angle ABC)$
$ \displaystyle {{y}^{2}}={{(5-x)}^{2}}+{{(4+x)}^{2}}-2(5-x)(4+x)\cos 120{}^\circ $
$ \displaystyle {{y}^{2}}=25-10x+{{x}^{2}}+16+8x+{{x}^{2}}+2(5-x)(4+x)\left( {\frac{1}{2}} \right)$
$ \displaystyle {{y}^{2}}=41-2x+2{{x}^{2}}-{{x}^{2}}+x+20$
$ \displaystyle {{y}^{2}}={{x}^{2}}-x+61$
$ \displaystyle \therefore {{y}^{2}}={{x}^{2}}-x+\frac{1}{4}+61-\frac{1}{4}$
$ \displaystyle \therefore {{y}^{2}}={{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75$
Since $ \displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}\ge 0\ $ for all $ \displaystyle x\in R$,
$ \displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75\ge 60.75$
$ \displaystyle \therefore {{y}^{2}}\ge 60.75$.
Therefore the minimum value of $\displaystyle {{y}^{2}}$ is $ \displaystyle 60.75$ and this value occurs when $ \displaystyle x=\frac{1}{2}$.
(a) Show that $ \displaystyle {{y}^{2}}={{x}^{2}}-x+61$.
(b) Find the minimum value of $ \displaystyle {{y}^{2}}$, and give the value of $ \displaystyle x$ for which this occurs.
Solution
$ \displaystyle BC=(4+x)$ cm,
$ \displaystyle \angle ABC=120{}^\circ $
$ \displaystyle AC=y$ cm.
(a) By the law of cosines,
$ \displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2\cdot AB\cdot AC\cos (\angle ABC)$
$ \displaystyle {{y}^{2}}={{(5-x)}^{2}}+{{(4+x)}^{2}}-2(5-x)(4+x)\cos 120{}^\circ $
$ \displaystyle {{y}^{2}}=25-10x+{{x}^{2}}+16+8x+{{x}^{2}}+2(5-x)(4+x)\left( {\frac{1}{2}} \right)$
$ \displaystyle {{y}^{2}}=41-2x+2{{x}^{2}}-{{x}^{2}}+x+20$
$ \displaystyle {{y}^{2}}={{x}^{2}}-x+61$
$ \displaystyle \therefore {{y}^{2}}={{x}^{2}}-x+\frac{1}{4}+61-\frac{1}{4}$
$ \displaystyle \therefore {{y}^{2}}={{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75$
Since $ \displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}\ge 0\ $ for all $ \displaystyle x\in R$,
$ \displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75\ge 60.75$
$ \displaystyle \therefore {{y}^{2}}\ge 60.75$.
Therefore the minimum value of $\displaystyle {{y}^{2}}$ is $ \displaystyle 60.75$ and this value occurs when $ \displaystyle x=\frac{1}{2}$.
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