## ccsoccerox 2 years ago Given the triangle below, what is sec ∡B? a. 4 sqrt33/33 b. 7/4 c. 4/7 d. 7 sqrt33/33

1. ccsoccerox

2. WATTAGWAN

Sec B = 1/ Cos B and cosB = A/7 so sec B = 7/A where A = $\sqrt{7^2 - 4^2}$

3. WATTAGWAN

Seriously? you can do the rest with a calculator.