Given:
The diagram of two right triangles.
To find:
The value of d and a.
Solution:
In a right angle triangle,
[tex]\tan \theta=\dfrac{Perpendicular}{Base}[/tex]
[tex]\sin \theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
(a) In first triangle,
[tex]\tan 60^\circ=\dfrac{d}{4}[/tex]
[tex]\sqrt{3}=\dfrac{d}{4}[/tex]
[tex]4\sqrt{3}=d[/tex]
Therefore, [tex]d=4\sqrt{3}[/tex].
(b) In second triangle,
[tex]\sin 45^\circ=\dfrac{a}{2}[/tex]
[tex]\dfrac{1}{\sqrt{2}}=\dfrac{a}{2}[/tex]
[tex]\dfrac{2}{\sqrt{2}}=a[/tex]
[tex]\sqrt{2}=a[/tex]
Therefore, [tex]a=\sqrt{2}[/tex].
