If in a triangle ABC, sinA = cosB then the value of cosC is?
A. $$\frac{{\sqrt 3 }}{2}$$
B. 0
C. 1
D. $$\frac{1}{{\sqrt 2 }}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\text{In }}\vartriangle {\text{, }}\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C}} = {\text{18}}{0^ \circ }\,....{\text{(i)}} \cr & {\text{sin A}} = {\text{cos B}} \cr & {\text{sin A}} = {\text{sin}}\left( {{{90}^ \circ } - {\text{B}}} \right){\text{ }} \cr & {\text{A}} = {90^ \circ } - {\text{B}} \cr & {\text{A}} + {\text{B}} = {90^ \circ }\,........(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \angle {\text{C}} = {90^ \circ } \cr & {\text{So, cos C }} = \cos {90^ \circ } = 0 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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