Solve the right triangle with the given sides and angles 17

Solve the right triangle with the given sides and angles. 1.7 cm 23) a = 2.1 cm, b Solve the problem. 24) The range r of a projectile is given by v2 sin 20, 32 r where v is the initial velocity and is the angle of elevation. If r is to be 800 ft and v-4500 ft/sec what must the angle of elevation be? Give your answer in degrees to the nearest hundredth.

Solution

23 Given : right angled triangle:

a = 2.1 cm , b = 1.7 cm

c = sqrt(a^2 +b^2) = sqrt(2.1^2 +1.7^2) = 2.70

a/sinA = c/sinC

sinA = a*sinC/c = 2.1*1/2.70

A = 51.05 deg

B = 180 -A -C = 38.95 deg

24) r = v^2/32 sin2theta

r = 8000 ft ; v = 4500ft/sec

sin2theta = 32*r/v^2 = 32*8000/(4500)^2 = 0.0126

2theta = sin^-1(0.0126)

=0.724

theta =0.36 degree (rounded off two places)