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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
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B
3817
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C
378
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D
3817
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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,

$$\cos \alpha =\dfrac{\sqrt {63}}{8}=\dfrac{3\sqrt 7}{8}$$


1110721_1049238_ans_a2bd1ca7277a4ae6b5f571685a941dec.PNG

Mathematics

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