Question-228105

Question Number 228105 by Math1 last updated on 23/Mar/26

Commented by mr W last updated on 24/Mar/26

$${NB}=\frac{{a}}{\mathrm{tan}\:\mathrm{16}°}\left(\mathrm{tan}\:\mathrm{53}°−\mathrm{tan}\:\mathrm{37}°\right)=\mathrm{2}{a} \\ $$

Commented by TonyCWX last updated on 24/Mar/26

My calculator suggested that ((tan(53°)−tan(37°))/(tan(16°))) = 2 though...

$$\mathrm{My}\:\mathrm{calculator}\:\mathrm{suggested}\:\mathrm{that}\:\frac{\mathrm{tan}\left(\mathrm{53}°\right)−\mathrm{tan}\left(\mathrm{37}°\right)}{\mathrm{tan}\left(\mathrm{16}°\right)}\:=\:\mathrm{2}\:\mathrm{though}… \\ $$

Commented by Ghisom_ last updated on 24/Mar/26

((tan x −tan (90°−x))/(tan (x−(90°−x))))=((tan x −cot x)/(−cot 2x))= =((((cos x)/(sin x))−((sin x)/(cos x)))/((1−2cos^2 x)/(2cos x sin x)))=2

$$\frac{\mathrm{tan}\:{x}\:−\mathrm{tan}\:\left(\mathrm{90}°−{x}\right)}{\mathrm{tan}\:\left({x}−\left(\mathrm{90}°−{x}\right)\right)}=\frac{\mathrm{tan}\:{x}\:−\mathrm{cot}\:{x}}{−\mathrm{cot}\:\mathrm{2}{x}}= \\ $$$$=\frac{\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}}{\frac{\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \:{x}}{\mathrm{2cos}\:{x}\:\mathrm{sin}\:{x}}}=\mathrm{2} \\ $$

Answered by TonyCWX last updated on 24/Mar/26

∠DAN = 90° − 37° − 2(16°) = 21° tan(16°) = (a/(AC)) ⇒ AC = (a/(tan(16°))) tan(37°) = ((a+DN)/(AC)) ⇒ AC = ((a+DN)/(tan(37°))) tan(53°) = ((a+DN+NB)/(AC)) ⇒ AC = ((a+DN+NB)/(tan(53°))) (a/(tan(16°))) = ((a+DN)/(tan(37°))) ⇒ DN = ((a[tan(37°)−tan(16°)])/(tan(16°))) (a/(tan(16°))) = ((a+a[((tan(37°)−tan(16°))/(tan(16°)))]+NB)/(tan(53°))) NB = ((a[tan(53°)−tan(37°)])/(tan(16°))) = ((atan(53°−37°)[1+tan(53°)tan(37°)])/(tan(16°))) = ((atan(16°)[1+1])/(tan(16°))) = 2a

$$\angle{DAN}\:=\:\mathrm{90}°\:−\:\mathrm{37}°\:−\:\mathrm{2}\left(\mathrm{16}°\right)\:=\:\mathrm{21}° \\ $$$$ \\ $$$$\mathrm{tan}\left(\mathrm{16}°\right)\:=\:\frac{{a}}{{AC}}\:\Rightarrow\:{AC}\:=\:\frac{{a}}{\mathrm{tan}\left(\mathrm{16}°\right)} \\ $$$$\mathrm{tan}\left(\mathrm{37}°\right)\:=\:\frac{{a}+{DN}}{{AC}}\:\Rightarrow\:{AC}\:=\:\frac{{a}+{DN}}{\mathrm{tan}\left(\mathrm{37}°\right)} \\ $$$$\mathrm{tan}\left(\mathrm{53}°\right)\:=\:\frac{{a}+{DN}+{NB}}{{AC}}\:\Rightarrow\:{AC}\:=\:\frac{{a}+{DN}+{NB}}{\mathrm{tan}\left(\mathrm{53}°\right)} \\ $$$$ \\ $$$$\frac{{a}}{\mathrm{tan}\left(\mathrm{16}°\right)}\:=\:\frac{{a}+{DN}}{\mathrm{tan}\left(\mathrm{37}°\right)}\:\Rightarrow\:{DN}\:=\:\frac{{a}\left[\mathrm{tan}\left(\mathrm{37}°\right)−\mathrm{tan}\left(\mathrm{16}°\right)\right]}{\mathrm{tan}\left(\mathrm{16}°\right)} \\ $$$$\frac{{a}}{\mathrm{tan}\left(\mathrm{16}°\right)}\:=\:\frac{{a}+{a}\left[\frac{\mathrm{tan}\left(\mathrm{37}°\right)−\mathrm{tan}\left(\mathrm{16}°\right)}{\mathrm{tan}\left(\mathrm{16}°\right)}\right]+{NB}}{\mathrm{tan}\left(\mathrm{53}°\right)} \\ $$$$ \\ $$$${NB}\:=\:\frac{{a}\left[\mathrm{tan}\left(\mathrm{53}°\right)−\mathrm{tan}\left(\mathrm{37}°\right)\right]}{\mathrm{tan}\left(\mathrm{16}°\right)}\: \\ $$$$=\:\frac{{a}\mathrm{tan}\left(\mathrm{53}°−\mathrm{37}°\right)\left[\mathrm{1}+\mathrm{tan}\left(\mathrm{53}°\right)\mathrm{tan}\left(\mathrm{37}°\right)\right]}{\mathrm{tan}\left(\mathrm{16}°\right)}\: \\ $$$$=\:\frac{{a}\mathrm{tan}\left(\mathrm{16}°\right)\left[\mathrm{1}+\mathrm{1}\right]}{\mathrm{tan}\left(\mathrm{16}°\right)}\: \\ $$$$=\:\mathrm{2}{a} \\ $$

Commented by TonyCWX last updated on 24/Mar/26

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