# Question 81ed9

Jun 13, 2016

#### Explanation:

i) $2 x \cos \left(\theta\right) = 4 \to \cos \left(\theta\right) = \frac{2}{x}$
ii) $\hat{C} = \arctan \left(\frac{\sqrt{15} x}{x}\right) = \arctan \left(\sqrt{15}\right)$
iii) $\sin \frac{\theta}{x} = \sin \frac{\hat{C}}{4}$ but $\sin \left(\hat{C}\right) = \frac{\sqrt{15} x}{4 x} = \frac{\sqrt{15}}{4}$

so $\sin \left(\theta\right) = \frac{\sqrt{15}}{16} x$
iv)# ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1 \to {\left(\frac{\sqrt{15}}{16} x\right)}^{2} + {\left(\frac{2}{x}\right)}^{2} = 1$

Jun 13, 2016

Steps as below

#### Explanation:

(i) For a triangle with sides $a , b \mathmr{and} c$ and angles $A , B \mathmr{and} C$ the cosine rule states:
${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$
Applying in $\Delta D B C$ where $\angle D = \theta$

${x}^{2} = {x}^{2} + {4}^{2} - 2 \times x \times 4 \cos \theta$, rearranging and solving for $\cos \theta$
$\implies 8 x \cos \theta = 16$
$\implies \cos \theta = \frac{16}{8 x}$
$\implies \cos \theta = \frac{2}{x}$, Proved.
(ii) In $\Delta A B C$
$\sin A \hat{C} B = \text{perpendicular"/"hypotenuse} = \frac{\sqrt{15} x}{4 x} = \frac{\sqrt{15}}{4}$
(iii) For a triangle with sides $a , b \mathmr{and} c$ and angles $A , B \mathmr{and} C$ the sine rule states:
$\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$
In $\Delta B C D$ applying the law of sines for angles $D \mathmr{and} C$ and using value from (ii) we get

$\sin \frac{\theta}{x} = \frac{\frac{\sqrt{15}}{4}}{4}$
$\sin \theta = \frac{\sqrt{15} x}{16}$
(iv) LHS $= {\sin}^{2} \theta + {\cos}^{2} \theta$ Inserting values as obtained above

$\implies L H S = {\left(\frac{\sqrt{15} x}{16}\right)}^{2} + {\left(\frac{2}{x}\right)}^{2}$
$= \frac{15 {x}^{2}}{256} + \frac{4}{x} ^ 2$, Setting it equal to RHS$= 1$ we obtain
$\frac{15 {x}^{2}}{256} + \frac{4}{x} ^ 2 = 1$
$\implies \frac{15 {x}^{2}}{256} + \frac{4}{x} ^ 2 - 1 = 0$
Rearranging and multiplying both sides with LCM of all the denominators, we obtain

$256 {x}^{2} \left[\frac{15 {x}^{2}}{256} - 1 + \frac{4}{x} ^ 2 = 0\right]$
$\implies 15 {x}^{4} - 256 {x}^{2} + 1024 = 0$, Proved