Given: t= 20;T =40 m=10 ; $\phi =20^{\circ}$

**Addendum height for each gear wheel**

We know that the pitch circle radius of the smaller gear wheel,

r =m.t/2 =(10×20)/2 =100 mm

and pitch circle radius of the larger gear wheel,

R=m.T/2 = (10×40)/2=200 mm.

$R_{A}$ =radius of addendum circle for the larger gear wheel , and

$r_{A}$ = radius of addendum circle for the smaller gear wheel.

Since the addendum on each wheel iis to be made of such a length that the line of contact on each side of the pitch point (i.e. the path of approach and the path of recess) has half the maximum possible length, therefore

Path of Approach, KP= 1/2 MP

or $\sqrt{(R_{A})^{2}-R^{2}\cos^{2}\phi}-R\sin\phi=\frac{r\sin\phi}{2}$

or $\sqrt{(R_{A})^{2}-200^{2}\cos^{2}20^{\circ}}-200\sin20^{\circ}=\frac{100\sin20^{\circ}}{2}=50 \sin 20^{\circ}$

$\sqrt{(R_{A})^{2}-35 320}=50 \sin 20^{\circ}+200 \sin 20^{\circ}=250\times 0.342=85.5$

$(R_{A})^{2}-35 320=85.5^{2}=7310$............(squaring both sides)

$(R_{A})^{2} =7310 + 35 320=42 630$ or $R_{A}=206.5 mm$

Addendum height for larger gear wheel,

= $R_{A} - R=206.5-200=6.5 mm$

Now, path of recess, PL=1/2 PN

or $\sqrt{(r_{A})^{2}-r^{2}\cos^{2}\phi}-r\sin\phi=\frac{R\sin\phi}{2}$

or $\sqrt{(r_{A})^{2}-100^{2}\cos^{2}20^{\circ}}-100\sin20^{\circ}=\frac{200\sin20^{\circ}}{2}=100 \sin 20^{\circ}$

$\sqrt{(R_{A})^{2}-100^{2}\cos^{2}20^{\circ}}=100 \sin 20^{\circ}+100 \sin 20^{\circ}=200\times 0.342=68.4$

$(r_{A})^{2}-8830=68.4^{2}=4680$............(squaring both sides)

$(R_{A})^{2} =4680 + 8830=13 510$ or $r_{A}=116.2 mm$

Addendum height for smaller gear wheel,

= $r_{A} - r=116.2-100=16.2 mm$

**Length of the path of contact**

We know that length of the path of contact

= KP+PL =1/2 MP + 1/2 PN =$\frac{(r+R)\sin\phi}{2}$

= $\frac{100+200}\sin 20^{\circ}{2}$=51.3 mm

**Contact Ratio**

We know that circular pitch

$p_{c}=\pi m =\pi\times 10=31.42 mm$

Contact ratio=Length of arc of contact / $p_{c}$ =54.6 /31.42 = 1.74 say.