Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 164426 by cortano1 last updated on 17/Jan/22

In ΔABC if { ((cot A.cot C = (1/2))),((cot B.cot C = (1/(18)))) :} then tan C = ?

$$\:\:{In}\:\Delta{ABC}\:{if}\:\begin{cases}{\mathrm{cot}\:{A}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{cot}\:{B}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{18}}}\end{cases} \\ $$$$\:{then}\:\mathrm{tan}\:\:{C}\:=\:? \\ $$

Commented by bobhans last updated on 17/Jan/22

let tan C=x>0 ⇒ (2/x)+((18)/x)+x = (2/x) .((18)/x). x ⇒ 20 +x^2 = 36⇒x=tan C = 4 ✓

$$\:\mathrm{let}\:\mathrm{tan}\:\mathrm{C}=\mathrm{x}>\mathrm{0}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{18}}{\mathrm{x}}+\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:.\frac{\mathrm{18}}{\mathrm{x}}.\:\mathrm{x} \\ $$$$\:\Rightarrow\:\mathrm{20}\:+\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{36}\Rightarrow\mathrm{x}=\mathrm{tan}\:\mathrm{C}\:=\:\mathrm{4}\:\checkmark \\ $$

Answered by som(math1967) last updated on 17/Jan/22

A+B+C=π C=π−(A+B) cotC=−cot(A+B) cotC= −((cotAcotB−1)/(cotB+cotA)) cotBcotC+cotAcotC=−cotAcotB+11 ⇒cotAcotB=1−(1/(18))−(1/2)=((18−1−9)/(18))=(8/(18))=(4/9) cot^2 Acot^2 Bcot^2 C=(1/2)×(1/(18))×(4/9)=(1/(81)) cotAcotBcotC=(1/9) tanC=(1/(cotC))=((cotAcotB)/(cotAcotBcotC))=(4/9)×(9/1) tanC=4

$${A}+{B}+{C}=\pi \\ $$$${C}=\pi−\left({A}+{B}\right) \\ $$$${cotC}=−{cot}\left({A}+{B}\right) \\ $$$${cotC}=\:−\frac{{cotAcotB}−\mathrm{1}}{{cotB}+{cotA}} \\ $$$${cotBcotC}+{cotAcotC}=−{cotAcotB}+\mathrm{11} \\ $$$$\Rightarrow{cotAcotB}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{18}−\mathrm{1}−\mathrm{9}}{\mathrm{18}}=\frac{\mathrm{8}}{\mathrm{18}}=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${cot}^{\mathrm{2}} {Acot}^{\mathrm{2}} {Bcot}^{\mathrm{2}} {C}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{18}}×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{81}} \\ $$$${cotAcotBcotC}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${tanC}=\frac{\mathrm{1}}{{cotC}}=\frac{{cotAcotB}}{{cotAcotBcotC}}=\frac{\mathrm{4}}{\mathrm{9}}×\frac{\mathrm{9}}{\mathrm{1}} \\ $$$${tanC}=\mathrm{4} \\ $$

Commented by cortano1 last updated on 17/Jan/22

only 4 sir

$${only}\:\:\mathrm{4}\:{sir} \\ $$

Commented by som(math1967) last updated on 17/Jan/22

yes tanC>0

$${yes}\:{tanC}>\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com