Answer:
[tex]\text{3. In case of this question, you may find the value of }v\text{ either by using a }\\\text{trigonometric ratio or use the concept of isosceles right triangle.}[/tex]
[tex]\textbf{Using trigonometric ratio:}\\\\\tan45^\circ=\dfrac{5\sqrt{2}}{v}\\\\\text{or, }1=\dfrac{5\sqrt{2}}{v}\\\\\text{or, }v=5\sqrt{2}[/tex]
[tex]\text{Remember, the opposite side of the angle taken as reference, in this case }\\45^\circ\text{ is the perpendicular}(p)\text{ and the remaining leg is the base}(b).\\\text{Here, we used tangent ratio because the side opposite to the reference angle}\\45^\circ\text{ was given already. But you may also use the cotangent ratio.}[/tex]
[tex]\textbf{Isosceles right triangle}\\\text{The two angles of the given triangle are }90^\circ\text{ and }45^\circ,\text{ so the remaining}\\\text{angle will be }45^\circ.\text{ So, the given triangle is isosceles so, the two sides }\\\text{opposite to the equal angles, }45^\circ\text{ and }45^\circ\text{ will also be equal.}[/tex]
[tex]\therefore\ v=5\sqrt2[/tex]
[tex]\text{Now, you may simply use the pythagoras theorem to calculate }w.\\w=\sqrt{v^2+\big(5\sqrt2\big)^2}=\sqrt{(5\sqrt{2})^2+50}=\sqrt{50+50}=10[/tex]
[tex]\text{Or you may use the trigonometric ratios again.}[/tex]
[tex]\text{4. Solution:}\\\\\sin45^\circ=\dfrac{x}{6}\\\\\text{or, }6\sin45^\circ=x\\\\\text{or, }x=\dfrac{6}{\sqrt2}\\\\\text{or, }x=\dfrac{6}{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=\dfrac{6\sqrt2}{2}\\\\\text{or, }x=3\sqrt2[/tex]
[tex]\text{Using the pythagoras theorem,}\\6^2=x^2+y^2\\\text{or, }36=(3\sqrt2)^2+y^2\\\text{or, }36=18+y^2\\\text{or, }y^2=18\\\text{or, }y=3\sqrt2[/tex]
Notice, this is again the right angled isosceles triangle because the remaining angle is 45 degrees, so x = y.
[tex]\text{5. Solution:}\\\\\sin60^\circ=\dfrac{x}{\dfrac{14\sqrt3}{3}}\\\\\text{or, }\dfrac{\sqrt3}{2}=\dfrac{3x}{14\sqrt3}\\\\\text{or, }14\sqrt3\times\sqrt3=6x\\\text{or, }6x=42\\\therefore\ x=7[/tex]
[tex]\text{Using the pythagoras theorem,}\\x^2+y^2=\bigg(\dfrac{14\sqrt3}{3}\bigg)^2\\\text{or, }7^2+y^2=\dfrac{196}{3}\\\\\text{or, }y^2=\dfrac{196}{3}-7^2=\dfrac{196}{3}-49\\\\\text{or, }y^2=\dfrac{196-147}{3}=\dfrac{49}{3}\\\\\text{or, }y=\dfrac{7}{\sqrt3}=\dfrac{7}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}\\\\\therefore\ y=\dfrac{7}{\sqrt3}[/tex]
[tex]\text{Or you may simply use a relevant trigonometric ratio.}\\\cos60^\circ=\dfrac{y}{\dfrac{14\sqrt3}{3}}\\\\\text{or, }\dfrac{1}{2}=\dfrac{y}{\dfrac{14\sqrt3}{3}}\\\\\text{or, }2y=\dfrac{14\sqrt3}{3}\\\\\text{or, }6y=14\sqrt3\\\\\text{or, }y=\dfrac{7\sqrt3}{3}[/tex]
