A)
(i) (ii) \[{{15}^{o}}\] \[{{45}^{o}}\]
B)
(i) (ii) \[{{17}^{o}}\] \[{{45}^{o}}\]
C)
(i) (ii) \[{{15}^{o}}\] \[{{33}^{o}}\]
D)
(i) (ii) \[{{17}^{o}}\] \[{{33}^{o}}\]
Correct Answer: D
Solution :
\[\angle CAQ+\angle QAP+PAB={{90}^{o}}\] \[\because \]\[AC\bot AB\] So, \[2x+{{7}^{o}}+{{40}^{o}}+x+{{4}^{o}}={{90}^{o}}\] \[\Rightarrow \]\[3x+51={{90}^{o}}\Rightarrow x={{13}^{o}}\] (i) \[\angle BAP=(x+{{4}^{o}})={{(13+4)}^{o}}={{17}^{o}}\] (ii)\[\angle CAQ=(2x+{{7}^{o}})={{(26+7)}^{o}}={{33}^{o}}\]
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