Question

Answer the question in the picture

Answer 1

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$