Question

# In the isosceles triangle ABC $$\left|\vec{AB}\right|=\left|\vec{BC}\right|=8$$, a point E divides AB internally in the ratio $$1:3$$, then the cosine of the angle between $$\vec{CE}$$ & $$\vec{CA}$$ is? (where $$\left|\vec{CA}\right|=12$$)

A
378
B
3817
C
378
D
3817

Solution

## The correct option is A $$\dfrac{3\sqrt{7}}{8}$$Consider the problem Given ,$$|AB|=|BC|=8$$Let,$$|\vec b|=8$$$$|\vec c|=12$$$$|\vec b-\vec c|=8$$$$\vec {CE}=\vec E-\vec C$$$$=\dfrac{\vec b}{4}-\vec c$$$$|\vec {CE}|=|\dfrac{\vec b}{4}-\vec c|=\sqrt {\dfrac{|\vec b|^2}{16}+|\vec c|^2-2.\dfrac{\vec b.\vec c}{4}}$$$$|\vec {CE}|=|\dfrac{\vec b}{4}-\vec c|=\sqrt {\dfrac{64}{16}+144-\dfrac{\vec b.\vec c}{2}}$$$$=\sqrt {148-\dfrac{\vec b.\vec c}{2}}$$   ----  (i)$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$$$$\cos A=\dfrac{8^2+12^2-8^2}{2.8.12}$$$$=\dfrac{3}{4}$$$$\vec b.\vec c=|\vec b||\vec c|\cos A$$$$=8 \times 12 \times \dfrac{3}{4}=72$$from (i)$$=\sqrt {148-\dfrac{72}{2}}=\sqrt {112}$$Now in triangle $$ABC$$  $$\dfrac{CE}{\sin A}=\dfrac{AE}{\sin \alpha }$$         $$(\cos A=\dfrac{3}{4},\;\;\sin A=\dfrac{\sqrt 7}{4})$$$$\dfrac{\sqrt {112}}{\dfrac{\sqrt7}{4}}=\dfrac{2}{\sin \alpha}$$$$\sin \alpha =\dfrac{2 \times \sqrt 7}{\sqrt {112}.4}$$$$=\dfrac{1}{8}$$then,$$\cos \alpha =\dfrac{\sqrt {63}}{8}=\dfrac{3\sqrt 7}{8}$$Mathematics

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