in triangle ABC, tan(angleCAB)=22/7 and the foot of perpendicular from A to BC is D if BD=3 DC=17 then find AD
Answers
Given :- in triangle ABC, tan(angleCAB) = 22/7 and the foot of perpendicular from A to BC is D if BD = 3, DC = 17 .
To Find :- Length of AD = ?
Solution :-
Let us assume that, length of AD is x .
now, in right ∆ADB, we have,
→ tan (∠DAB) = DB/AD
→ tan (∠DAB) = (17/x) ------ Eqn.(1)
and, in right ∆ADC, we have,
→ tan (∠DAC) = DC/AD
→ tan (∠DAC) = (3/x) ------ Eqn.(1)
now,
→ ∠CAB = ∠DAB + ∠DAC
using tan both sides ,
→ tan (∠CAB) = tan (∠DAB + ∠DAC)
using in RHS :-
- tan(A + B) = tan A + tan B / (1 - tan A * tan B)
putting values from Eqn.(1) and Eqn.(2) in RHS and given value in LHS now,
→ 22/7 = {(17/x) + (3/x)} / {1 - (17/x)(3/x)}
→ 22/7 = (20/x) / {1 - (51/x²)}
→ 22/7 = (20/x) * x²/(x² - 51)
→ 22/7 = 20x/(x² - 51)
→ 22(x² - 51) = 140x
→ 11x² - 70x - 561 = 0
→ 11x² - 121x + 51x - 561 = 0
→ 11x(x - 11) + 51(x - 11) = 0
→ (x - 11)(11x + 51) = 0
putting both equal to zero we get,
→ x = 11 and (-51/11)
since value of perpendicular height cant be in negative.
Hence, we can conclude that, length of AD(x) is equal to 11 cm .
Learn more :-
In ABC, AD is angle bisector,
angle BAC = 111 and AB+BD=AC find the value of angle ACB=?
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