Answer:
b. about 63.9 units and 41.0 units
Explanation:
In question ∠a= 29° and Side of a= 15 and b= 20
Using sine rule of congruence of triangle.
⇒
![(a)/(sin A) = (b)/(sin B) = (c)/(sin C)](https://img.qammunity.org/2020/formulas/mathematics/high-school/v8w98jpmxigoj9ahe3osq2n1ae1xr91rae.png)
⇒
![(15)/(Sin 29) = (20)/(sin B)](https://img.qammunity.org/2020/formulas/mathematics/high-school/y65g7pqito6q5bluldnoacdz3ij3a3ir2r.png)
Using value of sin 29°
⇒
![(15)/(0.49) = (20)/(sin B)](https://img.qammunity.org/2020/formulas/mathematics/high-school/iet883etvs0qcvua7h83xrfgluf6kvz9d2.png)
Cross multiplying both side.
⇒ Sin B=
![(20* 0.49)/(15) = 0.65](https://img.qammunity.org/2020/formulas/mathematics/high-school/9mhnb3o752s352syd0v0xvg6sthzo9r9p8.png)
∴ B= 41°
Now, we have the degree for ∠B= 41°.
Next, lets find the ∠C
∵ we know the sum total of angle of triangle is 180°
∴∠A+∠B+∠C= 180°
⇒
![29+41+B= 180](https://img.qammunity.org/2020/formulas/mathematics/high-school/pg5r610mnr6hwg0b2xs4pbwaij7ev901wp.png)
subtracting both side by 70°
∴∠C= 110°
Now, again using the sine rule to find the side of c.
![(b)/(SinB) = (c)/(SinC)](https://img.qammunity.org/2020/formulas/mathematics/high-school/5rfdr65ytkyn7c4evzkls47lqfxo1stude.png)
⇒
![(20)/(sin41) = (C)/(sin110)](https://img.qammunity.org/2020/formulas/mathematics/high-school/654ilb09jmgagetwp4z4cvszbk245gilp6.png)
Using the value of sine and cross multiplying both side.
⇒ C=
![(20* 0.94)/(0.65) = 28.92](https://img.qammunity.org/2020/formulas/mathematics/high-school/9wx6se9ucthuwriuea9q3see00a9yqw5os.png)
∴ Side C= 28.92.
Now, finding perimeter of angle of triangle
Perimeter of triangle= a+b+c
Perimeter of triangle=
![(15+20+28.92)= 63.9](https://img.qammunity.org/2020/formulas/mathematics/high-school/5cx3q83f0p6xvqe2c8ekowndm2tq8o7uri.png)
∴ Perimeter of triangle= 63.9 units