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1.     In \Delta \: ABC, right-angled at B, AB = 24 \: cm, BC = 7 \: cm. Determine :

             (i)\; \sin A, \cos A
             (ii)\; \sin C, \cos C

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We have,
In \Delta \: ABC\angleB = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm 
So, by using Pythagoras theorem,
\\AC^2 = AB^2 + BC^2\\ AC = \sqrt{AB^2+BC^2}
Therefore, AC = \sqrt{576+49}
                   AC = \sqrt{625}
                    AC = 25 cm

Now,
(i) \sin A = P/H = BC/AB = 7/25
    \cos A = B/H = BA/AC = 24/25

(ii)  For angle C, AB is perpendicular to the base (BC).  Here B  indicates to Base and P means perpendicular wrt angle \angleC
So,  \sin C = P/H = BA/AC = 24/25       
and  \cos C = B/H = BC/AC = 7/25
 

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manish

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