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Question-223157

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Question Number 223157 by cemoosky last updated on 16/Jul/25
Answered by som(math1967) last updated on 16/Jul/25
∠QPN+∠NML=55° ∴∠NPQ=55−40=15° ∠RQP=∠NPQ=15°
$$\angle{QPN}+\angle{NML}=\mathrm{55}° \\ $$$$\therefore\angle{NPQ}=\mathrm{55}−\mathrm{40}=\mathrm{15}° \\ $$$$\angle{RQP}=\angle{NPQ}=\mathrm{15}° \\ $$
Answered by mehdee7396 last updated on 16/Jul/25
γ=40⇒β=15⇒α=β=15
$$\gamma=\mathrm{40}\Rightarrow\beta=\mathrm{15}\Rightarrow\alpha=\beta=\mathrm{15} \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 16/Jul/25

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