In Δ PQR, side Q = 25, m < P = 13* and m < Q = 70*. Find side P to the nearest tenth of an integer

The answer to your question is p = 6.0 (real value 5.98)

Step-by-step explanation:

Data

q = 25

∠P = 13°

∠Q = 70°

p = ?

In this kind of problem, the length of the sides is written in lowercase letters and the angles in Upper case letters.

Process

1.- To solve this problem use the law of sines.

                 p/sin P = q/sin Q

-Solve for p

                 p = q sin P/ sin Q

- Substitution

                 p = 25 sin 13 / sin 70

-Simplification

                 p = 25(0.225) / (0.9397)

                 p = 5.6237 / 0.9397

- Result

                 p = 5.9846

Rounded to the nearest tenth

                 p = 6.0      

6.0

Step-by-step explanation:

25/sin(80) = p/sin(13)

p = sin(13) × 25/sin(80)

p = 5.984697798

the diameter is 16 cm. are you a teacher?

step-by-step explanation: