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In triangle ABC, we have angle A = 90^\circ and sin B = 5/7. Find cos B.

 Aug 2, 2022
 #1
avatar+2455 
0

Here's my attempt (I'm not good at trig): 

 

C

 

5             7

 

A                           B

 

Let \(AC = 5\) and \(BC = 7\)

 

By the Pythagorean Theorem, \(AB = \sqrt{24} = 2\sqrt6\)

 

So, \(\cos B = \color{brown}\boxed{2 \sqrt 6 \over 7}\)

 Aug 2, 2022
 #2
avatar+124598 
+2

Here's another way

 

cos B  =  sqrt (1 - sin^2B)  =  sqrt ( 1 - 25/49 )  =  sqrt ( 49 -25) / 7  =  sqrt ( 24)  / 7  =   

 

2sqrt (6)  /  7        (as BuilderBoi found !!!)

 

 

cool cool cool

 Aug 2, 2022
 #3
avatar+113 
+2

\(In\;the\; \Delta ABC, we\; have\; the \;\angle A\;=90^{\circ}.\;sin(B)\;=\; \frac{5}{7}.\;Find\;cos(B).\)

Sorry guys, I kinda f'ed up cos and sin in the drawing, but the calculation is correct.

 Aug 3, 2022
edited by tuffla2022  Aug 3, 2022
edited by tuffla2022  Aug 3, 2022
 #4
avatar+113 
+2

This is more like how it should look. Sorry, guys.

tuffla2022  Aug 3, 2022

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