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In a triangle PQR, P is the largest angle and cos p = 1/3. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
- a)16
- b)18
- c)24
- d)22
Correct answer is option 'B,D'. Can you explain this answer?
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Verified Answer
In a triangle PQR, P is the largest angle and cos p = 1/3.Further the ...
Let PN = x, QL = x + 2, RM = x + 4 where x is an even integer.
Then PM = PN = x, QN = QL = x + 2 and RL = RM = x + 4
So that PQ = 2x + 2, QR = 2x + 6, PR = 2x + 4
Most Upvoted Answer
In a triangle PQR, P is the largest angle and cos p = 1/3.Further the ...
Given information:
- In triangle PQR, angle P is the largest angle and cos P = 1/3.
- The incircle of the triangle touches the sides PQ, QR, and RP at points N, L, and M respectively.
- The lengths of PN, QL, and RM are consecutive even integers.
To find:
Possible length(s) of the side(s) of the triangle.
Solution:
Step 1: Finding the angles of the triangle
Since cos P = 1/3, we can use the inverse cosine function to find angle P:
P = cos^(-1)(1/3) ≈ 70.53°
Since angle P is the largest angle, we know that angles Q and R are smaller than 70.53°.
Step 2: Understanding the properties of the incircle
The incircle of a triangle is the circle that is tangent to all three sides of the triangle.
The lengths of the segments PN, QL, and RM are the distances from the vertices of the triangle to the points where the incircle touches the sides.
Step 3: Establishing properties of the consecutive even integers
Let the consecutive even integers be 2n, 2n+2, and 2n+4.
Step 4: Applying properties of the incircle
According to the properties of the incircle, the lengths of the segments PN, QL, and RM are given by:
PN = QL = RM = r, where r is the radius of the incircle.
We know that the radius of the incircle is given by the formula:
r = (s - a)tan(A/2),
where s is the semi-perimeter of the triangle, a is the length of the side opposite to angle A.
Step 5: Finding the possible lengths of the sides
Using the formula for the radius of the incircle, we can find the possible lengths of the sides of the triangle by substituting the values of the angles and the consecutive even integers.
For the first set of consecutive even integers (2n, 2n+2, 2n+4), we find:
s = (2n + 2n+2 + 2n+4)/2 = 6n + 6,
a = 2n,
A = 70.53°.
Substituting these values into the formula for the radius, we get:
r = (6n + 6 - 2n)tan(70.53°/2) = 4n(tan(35.26°)).
Since r must be an even integer, we can solve for n by trial and error.
Let's try n = 3:
r = 4(3)(tan(35.26°)) ≈ 8.36,
which is not an even integer.
Let's try n = 4:
r = 4(4)(tan(35.26°)) ≈ 11.15,
which is not an even integer.
Let's try n = 5:
r = 4(5)(tan(35.26°)) ≈ 13.94,
which is not an even integer.
Let's try n = 6:
r = 4
- In triangle PQR, angle P is the largest angle and cos P = 1/3.
- The incircle of the triangle touches the sides PQ, QR, and RP at points N, L, and M respectively.
- The lengths of PN, QL, and RM are consecutive even integers.
To find:
Possible length(s) of the side(s) of the triangle.
Solution:
Step 1: Finding the angles of the triangle
Since cos P = 1/3, we can use the inverse cosine function to find angle P:
P = cos^(-1)(1/3) ≈ 70.53°
Since angle P is the largest angle, we know that angles Q and R are smaller than 70.53°.
Step 2: Understanding the properties of the incircle
The incircle of a triangle is the circle that is tangent to all three sides of the triangle.
The lengths of the segments PN, QL, and RM are the distances from the vertices of the triangle to the points where the incircle touches the sides.
Step 3: Establishing properties of the consecutive even integers
Let the consecutive even integers be 2n, 2n+2, and 2n+4.
Step 4: Applying properties of the incircle
According to the properties of the incircle, the lengths of the segments PN, QL, and RM are given by:
PN = QL = RM = r, where r is the radius of the incircle.
We know that the radius of the incircle is given by the formula:
r = (s - a)tan(A/2),
where s is the semi-perimeter of the triangle, a is the length of the side opposite to angle A.
Step 5: Finding the possible lengths of the sides
Using the formula for the radius of the incircle, we can find the possible lengths of the sides of the triangle by substituting the values of the angles and the consecutive even integers.
For the first set of consecutive even integers (2n, 2n+2, 2n+4), we find:
s = (2n + 2n+2 + 2n+4)/2 = 6n + 6,
a = 2n,
A = 70.53°.
Substituting these values into the formula for the radius, we get:
r = (6n + 6 - 2n)tan(70.53°/2) = 4n(tan(35.26°)).
Since r must be an even integer, we can solve for n by trial and error.
Let's try n = 3:
r = 4(3)(tan(35.26°)) ≈ 8.36,
which is not an even integer.
Let's try n = 4:
r = 4(4)(tan(35.26°)) ≈ 11.15,
which is not an even integer.
Let's try n = 5:
r = 4(5)(tan(35.26°)) ≈ 13.94,
which is not an even integer.
Let's try n = 6:
r = 4
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In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer?
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In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer?.
In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer?.
Solutions for In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer?, a detailed solution for In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? has been provided alongside types of In a triangle PQR, P is the largest angle and cos p = 1/3.Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)a)16b)18c)24d)22Correct answer is option 'B,D'. Can you explain this answer? theory, EduRev gives you an
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