Question-219949

Question Number 219949 by Spillover last updated on 04/May/25

Answered by vnm last updated on 04/May/25

Let AH=1 be the altitude of triangle ABC ∡BAH=60°, ∡MAH=45°−y ∡CAH=75°, ∡NAH=45°+y tan 60°−tan (45°−y)=tan 75°−tan (45°+y) y=22.5° x=∡BAH−(45°−y)=37.5°

$$\mathrm{Let}\:{AH}=\mathrm{1}\:\mathrm{be}\:\mathrm{the}\:\mathrm{altitude}\:\mathrm{of}\: \\ $$$$\mathrm{triangle}\:{ABC} \\ $$$$\measuredangle{BAH}=\mathrm{60}°,\:\measuredangle{MAH}=\mathrm{45}°−{y} \\ $$$$\measuredangle{CAH}=\mathrm{75}°,\:\measuredangle{NAH}=\mathrm{45}°+{y} \\ $$$$\mathrm{tan}\:\mathrm{60}°−\mathrm{tan}\:\left(\mathrm{45}°−{y}\right)=\mathrm{tan}\:\mathrm{75}°−\mathrm{tan}\:\left(\mathrm{45}°+{y}\right) \\ $$$${y}=\mathrm{22}.\mathrm{5}° \\ $$$${x}=\measuredangle{BAH}−\left(\mathrm{45}°−{y}\right)=\mathrm{37}.\mathrm{5}° \\ $$

Commented by Spillover last updated on 04/May/25