Problem of the Week 2 Update
Two correct solutions were received. Congratulations to Mr. Isaiah Duazo an ECE student of Rizal Technological University and Mr. Jobert Gimeno an ECE student of PUP Sta. Mesa for submitting the correct solutions.
Solution by Isaiah Duazo:
tan A = 22/7
let A is equal to B + D
tan A = TAn (B + D)
(1) Tan (B + D) = (Tan B + Tan D)/[1 - (TAn B)(Tan D)]
Let the length of the altitude be h:
Tan B = 3/h
TAn D = 17/h
Substituting the value of Tan A, Tan B, and Tan D to Equation 1 and solving for h:
h = 11 units
Area of triangle = (base)(height)/2 = (20)(11)/2 = 110 square units
Solution by Isaiah Duazo:
tan A = 22/7
let A is equal to B + D
tan A = TAn (B + D)
(1) Tan (B + D) = (Tan B + Tan D)/[1 - (TAn B)(Tan D)]
Let the length of the altitude be h:
Tan B = 3/h
TAn D = 17/h
Substituting the value of Tan A, Tan B, and Tan D to Equation 1 and solving for h:
h = 11 units
Area of triangle = (base)(height)/2 = (20)(11)/2 = 110 square units
posted by zarcsman at
5:33 PM
2 Comments:
Sir Zarco. . .
nakuha ko na p0h ung answer. . .
tanA = 22/7
let x: be the base angle of the side whose length is 17.
let y: be the base angle of the side whose length is 3.
Let s be the perpendicular side :
therefore: x+y=A
tanx=17/s ;tany=3/s
s=17/tanx;s=3/tany
17/tanx=3/tany
3tanx=17tany
tanx=17tany/3
tanA=tan(x+y)
22/7=tanx+tany/1-tanxtany
22(1-tanxtany)=7(tanx+tany)
22-22tanxtany=7tanx+7tany
Since tanx =17tany/3,
22-22(17tany/3)tany=7(17tany/3)+ 7tany
22-(374tan²y/3)=(119tany/3)+7tany
66-374tan²y=119tany+21tany
-374tan²y-140tany+66=0
374tan²y+140tany-66=0
Using quadratic formula :
tany =3/11 ; -11/17
Using 3/11=tany,y=15.2551187...deg
tanx=17tany/3
tanx=17(3/11)/3
tanx=17/11
x=57.09475708
s=17/tanx
s=17/(17/11) ; s= 11
Area = ½bh
= ½(20)(11)
=10(11)
=110sq. units.
hehhe answer by: Libres Emmanuel
From:PUP Taguig Campus
10:37 PM
By
Anonymous, at 10:47 PM
just click the link below
for my answer
http://img60.imageshack.us/my.php?image=week2500210uw2.jpg
By
Anonymous, at 7:46 AM
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