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Problem related to Incircle of a triangle
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Given a $triangleABC$, Draw an incircle touching sides $AB,BC,AC$ at points $D,E,F$ respectively.Given $angleA=60°$, and lengths of $AD=5$ cm and $DB=3$cm.
Calculate the side length of $BC$.
My Attempt:
I know that AD=AF, CF=CE and BD=BE (Tangents to a circle from a common point)
I can also write,
$$AD=rcotfracA2$$
$$BE=rcotfracB2$$
$$CF=rcotfracC2$$
Putting AD=$5$ cm, and $angleA=60°$.
I calculated $$r=frac5sqrt3$$
Then calculated $$cotfracB2=frac3sqrt35$$
How do I proceed next?
I have in mind to use something like, $$cotfracA2+cotfracB2+cotfracC2=cotfracA2cotfracB2cotfracC2$$
Am I solving this right?
trigonometry triangle
asked Jul 15 at 13:39
prog_SAHIL
850217
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up vote
0
down vote
favorite
Given a $triangleABC$, Draw an incircle touching sides $AB,BC,AC$ at points $D,E,F$ respectively.Given $angleA=60°$, and lengths of $AD=5$ cm and $DB=3$cm.
Calculate the side length of $BC$.
My Attempt:
I know that AD=AF, CF=CE and BD=BE (Tangents to a circle from a common point)
I can also write,
$$AD=rcotfracA2$$
$$BE=rcotfracB2$$
$$CF=rcotfracC2$$
Putting AD=$5$ cm, and $angleA=60°$.
I calculated $$r=frac5sqrt3$$
Then calculated $$cotfracB2=frac3sqrt35$$
How do I proceed next?
I have in mind to use something like, $$cotfracA2+cotfracB2+cotfracC2=cotfracA2cotfracB2cotfracC2$$
Am I solving this right?
trigonometry triangle
asked Jul 15 at 13:39
prog_SAHIL
850217
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a $triangleABC$, Draw an incircle touching sides $AB,BC,AC$ at points $D,E,F$ respectively.Given $angleA=60°$, and lengths of $AD=5$ cm and $DB=3$cm.
Calculate the side length of $BC$.
My Attempt:
I know that AD=AF, CF=CE and BD=BE (Tangents to a circle from a common point)
I can also write,
$$AD=rcotfracA2$$
$$BE=rcotfracB2$$
$$CF=rcotfracC2$$
Putting AD=$5$ cm, and $angleA=60°$.
I calculated $$r=frac5sqrt3$$
Then calculated $$cotfracB2=frac3sqrt35$$
How do I proceed next?
I have in mind to use something like, $$cotfracA2+cotfracB2+cotfracC2=cotfracA2cotfracB2cotfracC2$$
Am I solving this right?
trigonometry triangle
asked Jul 15 at 13:39
prog_SAHIL
850217
Given a $triangleABC$, Draw an incircle touching sides $AB,BC,AC$ at points $D,E,F$ respectively.Given $angleA=60°$, and lengths of $AD=5$ cm and $DB=3$cm.
Calculate the side length of $BC$.
My Attempt:
I know that AD=AF, CF=CE and BD=BE (Tangents to a circle from a common point)
I can also write,
$$AD=rcotfracA2$$
$$BE=rcotfracB2$$
$$CF=rcotfracC2$$
Putting AD=$5$ cm, and $angleA=60°$.
I calculated $$r=frac5sqrt3$$
Then calculated $$cotfracB2=frac3sqrt35$$
How do I proceed next?
I have in mind to use something like, $$cotfracA2+cotfracB2+cotfracC2=cotfracA2cotfracB2cotfracC2$$
Am I solving this right?
trigonometry triangle
asked Jul 15 at 13:39
prog_SAHIL
850217
edited Jul 15 at 14:20
asked Jul 15 at 13:39
prog_SAHIL
850217
asked Jul 15 at 13:39
prog_SAHIL
850217
asked Jul 15 at 13:39
prog_SAHIL
850217
850217
add a comment |Â
add a comment |Â
2 Answers
2
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up vote
1
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accepted
Hint: Use that
$$a^2=8^2+(2+a)^2-2cdot 8cdot (a+2)cos(60^circ)$$
solve this equation for $a$
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
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up vote
1
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Suppose the incenter is $P$. Then $r=DP$.
$$requirecancelr=tan 30^circcdot5=frac5sqrt33$$
Then:
$$angle ABP=arctanleft(frac5sqrt39right)impliesangle B=2angle ABP=2arctanleft(frac5sqrt39right)\
angle C=120^circ-2arctanleft(frac5sqrt39right)approx 32.20^circ\
fracsin ABC=fracsin CABimplies BC=frac(sin A)cdot ABsin C=frac4sqrt3sinleft(120^circ-2arctanleft(frac5sqrt39right)right)=fraccancel4sqrt3fraccancel4 sqrt313=13$$
answered Jul 15 at 14:09
John Glenn
1,659223
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: Use that
$$a^2=8^2+(2+a)^2-2cdot 8cdot (a+2)cos(60^circ)$$
solve this equation for $a$
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
1
down vote
accepted
Hint: Use that
$$a^2=8^2+(2+a)^2-2cdot 8cdot (a+2)cos(60^circ)$$
solve this equation for $a$
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: Use that
$$a^2=8^2+(2+a)^2-2cdot 8cdot (a+2)cos(60^circ)$$
solve this equation for $a$
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
Hint: Use that
$$a^2=8^2+(2+a)^2-2cdot 8cdot (a+2)cos(60^circ)$$
solve this equation for $a$
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
answered Jul 15 at 13:59
Dr. Sonnhard Graubner
66.9k32659
66.9k32659
add a comment |Â
add a comment |Â
up vote
1
down vote
Suppose the incenter is $P$. Then $r=DP$.
$$requirecancelr=tan 30^circcdot5=frac5sqrt33$$
Then:
$$angle ABP=arctanleft(frac5sqrt39right)impliesangle B=2angle ABP=2arctanleft(frac5sqrt39right)\
angle C=120^circ-2arctanleft(frac5sqrt39right)approx 32.20^circ\
fracsin ABC=fracsin CABimplies BC=frac(sin A)cdot ABsin C=frac4sqrt3sinleft(120^circ-2arctanleft(frac5sqrt39right)right)=fraccancel4sqrt3fraccancel4 sqrt313=13$$
answered Jul 15 at 14:09
John Glenn
1,659223
add a comment |Â
up vote
1
down vote
Suppose the incenter is $P$. Then $r=DP$.
$$requirecancelr=tan 30^circcdot5=frac5sqrt33$$
Then:
$$angle ABP=arctanleft(frac5sqrt39right)impliesangle B=2angle ABP=2arctanleft(frac5sqrt39right)\
angle C=120^circ-2arctanleft(frac5sqrt39right)approx 32.20^circ\
fracsin ABC=fracsin CABimplies BC=frac(sin A)cdot ABsin C=frac4sqrt3sinleft(120^circ-2arctanleft(frac5sqrt39right)right)=fraccancel4sqrt3fraccancel4 sqrt313=13$$
answered Jul 15 at 14:09
John Glenn
1,659223
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose the incenter is $P$. Then $r=DP$.
$$requirecancelr=tan 30^circcdot5=frac5sqrt33$$
Then:
$$angle ABP=arctanleft(frac5sqrt39right)impliesangle B=2angle ABP=2arctanleft(frac5sqrt39right)\
angle C=120^circ-2arctanleft(frac5sqrt39right)approx 32.20^circ\
fracsin ABC=fracsin CABimplies BC=frac(sin A)cdot ABsin C=frac4sqrt3sinleft(120^circ-2arctanleft(frac5sqrt39right)right)=fraccancel4sqrt3fraccancel4 sqrt313=13$$
answered Jul 15 at 14:09
John Glenn
1,659223
Suppose the incenter is $P$. Then $r=DP$.
$$requirecancelr=tan 30^circcdot5=frac5sqrt33$$
Then:
$$angle ABP=arctanleft(frac5sqrt39right)impliesangle B=2angle ABP=2arctanleft(frac5sqrt39right)\
angle C=120^circ-2arctanleft(frac5sqrt39right)approx 32.20^circ\
fracsin ABC=fracsin CABimplies BC=frac(sin A)cdot ABsin C=frac4sqrt3sinleft(120^circ-2arctanleft(frac5sqrt39right)right)=fraccancel4sqrt3fraccancel4 sqrt313=13$$
answered Jul 15 at 14:09
John Glenn
1,659223
answered Jul 15 at 14:09
John Glenn
1,659223
answered Jul 15 at 14:09
John Glenn
1,659223
answered Jul 15 at 14:09
John Glenn
1,659223
1,659223
add a comment |Â
add a comment |Â
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