Solutions of Triangles - Key Concepts

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Questions and Answers

In triangle ABC, if the medians AD and BE have lengths of 4 and ∠DAB = π/6 while ∠ABE = π/3, what is the area of triangle ABC?

The area of triangle ABC is $32/√3$.

If cos²(A/2) + c cos²(A/2) = 3b/2 in triangle ABC, what relationship do the sides a, b, and c satisfy?

The sides satisfy $a + b = c$.

Given the sides of a triangle as sina, cosa, and √(1 + sina cosa) for $0 < a < π/2$, what is the greatest angle of the triangle?

The greatest angle of the triangle is $90°$.

In a triangle ABC with ∠C = π/2, what does 2(r + R) equal, where r is the inradius and R the circumradius?

It equals $a + b$. Signup and view all the answers

If the altitudes from vertices A, B, C of triangle ABC are in H.P., what relationship do sinA, sinB, sinC have?

sinA, sinB, sinC are in G.P. Signup and view all the answers

In trapezium ABCD with AB and CD parallel and BC perpendicular to CD, how do you express AB if BC = p and CD = q?

AB is given by $(p^2 + q^2) rac{ ext{sine}}{p imes ext{cose} + q imes ext{sine}}$. Signup and view all the answers

Given the coordinates of the orthocenter A(-3,5) and centroid B(3,3) of triangle ABC, what is the radius of the circle having segment AC as diameter?

The radius is $3√5 / 2$. Signup and view all the answers

If a triangle has sides a = 6, b = 3, and cos(A - B) = 4/5, how would you find its area?

The area can be calculated using the formula involving sides and cosine of the angle difference. Signup and view all the answers

What is the relationship between the angles A, B, and C if they are in an arithmetic progression?

The angles A, B, and C can be expressed as A = B - d, B, and C = B + d for some value d. Signup and view all the answers

If the lengths of the sides opposite to A, B, and C are a, b, and c respectively, what can be said about the area of triangle ABC when given the area as 15√3?

The area can be calculated using Heron's formula or by using the formula: area = 1/2 * base * height. Signup and view all the answers

Determine the possible values of x for which the sides of the triangle satisfy the equations a = x² + x + 1 and b = x² - 1.

The possible values of x are those which satisfy both equations and maintain the triangle inequality. Signup and view all the answers

For triangle PQR where cos P = 1/3, how does this affect the measurement of angle P?

This situation indicates that angle P is obtuse since cos P is positive, leading to an angle greater than 90 degrees. Signup and view all the answers

In triangle XYZ, what does the condition (s-x)/4 = (s-y)/3 = (s-z)/2 suggest about the relative sizes of the sides?

This condition implies a specific proportional relationship among the sides of the triangle in relation to the semi-perimeter 's'. Signup and view all the answers

What is the significance of the radius of the incircle in determining the area of triangle ABC?

The area of triangle ABC can be directly computed as A = r * s, where r is the radius of the incircle and s is the semi-perimeter. Signup and view all the answers

Determine the meaning of the median meeting the side at S in terms of triangle PQR.

The median from R to side PQ means that segment RS divides PQ into two equal lengths, contributing to centroid calculations. Signup and view all the answers

What can be concluded about the lengths of sides in triangle ABC if sides a, b, and c are stated and area is given?

The lengths must satisfy the triangle inequality theorem, ensuring that the sum of any two sides is greater than the third. Signup and view all the answers

If 'r' is the inradius and 'R' is the circumradius of a triangle, what is the expression for 2(r + R)?

2(r + R) = a + b + c. Signup and view all the answers

What does the equation $2ac \sin(A - B + C)$ equal in terms of the side lengths of triangle ABC?

$2ac \sin(A - B + C) = a^2 + b^2 - c^2$. Signup and view all the answers

If the incentre 'I' of triangle ABC has an inradius 'r', and D, E, F are the feet of the perpendiculars from I, how is the relation $\frac{r}{r - r_1} + \frac{r}{r - r_2} + \frac{r}{r - r_3}$ expressed?

$\frac{r}{r - r_1} + \frac{r}{r - r_2} + \frac{r}{r - r_3} = \frac{(r - r_1)(r - r_2)(r - r_3)}{r_1 r_2 r_3}$. Signup and view all the answers

What inequality relates the area $A$ of triangle ABC with its side lengths a, b, c?

A ≤ $\sqrt{(a + b + c)abc}$. Signup and view all the answers

Under what condition does equality occur in the inequality $A ≤ \sqrt{(a+b+c)abc}$?

Equality occurs if and only if a = b = c. Signup and view all the answers

Which of the following does NOT uniquely determine an acute-angled triangle ABC: a, sinA, sinB or a, b, c?

A: a, sinA, sinB does NOT uniquely determine the triangle. Signup and view all the answers

What is the ratio of angles A, B, C for a triangle where the side lengths ratio is 1:√3:2?

The ratio A:B:C is 3:5:2. Signup and view all the answers

What is the area of an isosceles triangle with an angle of 120° and an inradius of √3?

The area is 12 + 7√3 sq. units. Signup and view all the answers

Given the inradii of triangles ABM and ACM as r₁ and r₂, what could be the possible range of r₁/r₂?

The range of r₁/r₂ depends on the specific dimensions of triangles ABM and ACM, typically $r_1/r_2 > 0$. Signup and view all the answers

How can you prove that 1/P₁ + 1/P₂ + 1/P₃ = 1/r + 1/r₁ + 1/r₂ + 1/r₃ for a triangle?

This can be proved using the relation between the area and the altitudes of the triangle, linking them through the inradius. Signup and view all the answers

In the triangle ABC, how does bc/r₁ + ca/r₂ + ab/r₃ = 2R[(b+c)/a + (c+a)/b + (a+b)/c - 3] get established?

This identity is derived from the relationship between the sides of the triangle and the respective inradii, incorporating the circumradius R. Signup and view all the answers

If (b+c)/11 = (c+a)/12 = (a+b)/13 in triangle ABC, how can you conclude cos A/7 = cos B/19 = cos C/25?

This can be shown by using the law of cosines and the relationships between the sides and angles of triangle ABC. Signup and view all the answers

For triangle ABC where B = 3C, what is the expression for cosC and sinC?

cosC = (b-c)/(2c) * , \sqrt{3} : and : sinC = \frac{\sqrt{3}}{b+c} * \sqrt{3}. Signup and view all the answers

How can you prove that ∠ABC = π/2 given the median BD = √3/4 * (b+c) / 2c in triangle ABC?

By leveraging the properties of medians and applying the Pythagorean theorem, the angle can be established. Signup and view all the answers

In a rhombus ABCD with circumradii of triangles ABD & ACD being 12.5 and 25 respectively, how do you find the area of the rhombus?

The area of the rhombus can be calculated using the formula Area = 1/2 * d₁ * d₂, where d₁ and d₂ are the diagonals. Signup and view all the answers

In triangle ABC, if a² + b² = 101c², how can you express cotC / (cotA + cotB)?

This can be expressed as cotC = 50.5(cotA + cotB), emphasizing the relationship among the cotangents and sides. Signup and view all the answers

If $2a²b² + 2b²c² = a² + b² + c²$ in triangle ABC, what is the measure of angle B?

π/4 Signup and view all the answers

Given an area of a triangle as $10√3$ sq.cm and perimeter as 20 cm with ∠C = 60°, find side c.

8 Signup and view all the answers

In triangle ABC with points D and E on side BC, if $rac{sin(x+y)sin(y+z)}{sinx sinz}$ equals a value, what is that value?

4 Signup and view all the answers

If in triangle ABC, $a = 5$, $b = 4$, and $cos(A - B) = rac{31}{32}$, what is side c?

6 Signup and view all the answers

For a triangle with sides in the ratio 3:7:8, what is the ratio of R to r?

5:2 Signup and view all the answers

What is the result of $rac{(b²sin²C + c²sin²B)}{Δ}$ in triangle ABC?

2 Signup and view all the answers

If $1/r1 + 1/r2 + 1/r3 = 3$, what does $ an(tanB/2 + tanC/2)$ equal?

1 Signup and view all the answers

In triangle ABC, if $cos²A + cos²B - cos²C = 1$, what type of triangle is it?

right-angled Signup and view all the answers

In a triangle ABC, if b² + c² = 3a², prove that cotB + cotC - cotA = 0.

Using the Law of Cosines, we can write: cos B = (a² + c² - b²)/(2ac) and cos C = (a² + b² - c²)/(2ab). Substituting these values and b² + c² = 3a² into the expression cotB + cotC - cotA, we get: cotB + cotC - cotA = (a² + c² - b²)/(2abc) + (a² + b² - c²)/(2abc) - (b² + c² - a²)/(2abc) = (3a² - b² - c²)/(2abc) = 0. Signup and view all the answers

Prove that, in a triangle ABC, 2sinAcosB = sinC if cosA/a = cosB/b = cosC/c.

From the given condition, we have cosB = (b/a)cosA and cosC = (c/a)cosA. Applying the Sine Rule, we obtain sinA/a = sinB/b = sinC/c. Now substituting cosB and cosC in 2sinAcosB = sinC, we get: 2sinA[(b/a)cosA] = (c/a)sinA. Simplifying, we get bsinAcosA = csinA. From Sine Rule, sinA/a = sinB/b = sinC/c, therefore: sinBcosA = sinC, which is true due to the sine rule, thus proving the statement. Signup and view all the answers

If a1, a2, a3, ... is a geometric sequence and b1, b2, b3, ... is a geometric sequence with b1 = 1, b2 = √7 - √28 + 1, a1 = √28, and Σn=1 ∞1/an = Σn=1 ∞ bn, find the area of the triangle with side lengths a1, a2, and a3. Express the answer in the form p/q, where p and q are relatively prime.

Firstly, calculate b2 = √7 - √28 + 1 = 1 - √7. The common ratio of the sequence b is (1 - √7)/1 = 1 - √7. Since Σn=1 ∞1/an = Σn=1 ∞ bn, we have 1/(a1(1 - r)) = b1/(1 - r) where r is the common ratio of the geometric sequence a. Plugging in the values, we get: 1/(√28(1 - r)) = 1/(1 - (1 - √7)). Solving, we find r = 1/√7. Therefore, a2 = a1 * r = √28 * ( 1/√7) = 2 and a3 = a2 * r = 2 * (1/√7) = 2/√7. To find the area, we can use Heron's formula: K = √(s(s-a1)(s-a2)(s-a3)), where s = (a1 + a2 + a3)/2 = (√28 + 2 + 2/√7)/2. Solving this, we get the area as 2/√7. Therefore, p + q = 2 + 7 = 9. Signup and view all the answers

In a triangle ABC, if a tan A + b tan B = (a + b) tan (A+ B)/2, prove that triangle ABC is isosceles.

We know that tan (A + B) = (tanA + tanB)/(1 - tanAtanB). Substituting this into the given equation, we get: a tanA + b tanB = (a + b)(tanA + tanB)/(2(1 - tanAtanB)). Simplifying and rearranging, we get: (a + b) tanA tanB = (a - b)(tanA + tanB). This can be further simplified to: 2btanA tanB = (a-b)tanA + (a-b)tanB. Since tanA and tanB are non-zero (as A and B are angles in a triangle), we can divide both sides by tanA tanB, giving: 2b = (a - b)/tanB + (a - b)/tanA. Using the identity cot x = 1/tan x, we can rewrite this as: 2b = (a - b) cotB + (a - b) cotA. Now, using the formula cot(A/2) = (s - a)/r, where s is the semi-perimeter of the triangle and r is the inradius, we get: 2b = (a - b)[(s - b)/r + (s - a)/r]. Simplifying, we get: 2b = (a - b)(2s - a - b)/r. Since r is non-zero, we must have 2b = (a - b)(2s - a - b). Expanding and simplifying: 2b = (a - b)(a + c). Since a, b, and c are side lengths, we must have a = b or c = a. Therefore, triangle ABC is isosceles. Signup and view all the answers

In a triangle ABC, if loga² = logb² + logc² - log(2bccosA), what can you say about the triangle ABC?

Using the properties of logarithms, we can simplify the given equation: loga² = log(b²c²/(2bccosA)). This simplifies further to: loga² = log(bc/(2cosA)). Since a², bc/(2cosA) > 0, we can remove the log function to get: a² = bc/(2cosA). Using the Law of Cosines, we know that a² = b² + c² - 2bccosA. Substituting this into the previous equation, we get: b² + c² - 2bccosA = bc/(2cosA). Simplifying and multiplying both sides by 2cosA, we get: 2b²cosA + 2c²cosA - 4bccos²A = bc. Rearranging the terms, we obtain: 2b²cosA + 2c²cosA - bc - 4bccos²A = 0. Factoring by grouping, we get: (2cosA - 1)(b² + c² - 2bccosA) = 0. Since b² + c² - 2bccosA = a² > 0, we must have 2cosA - 1 = 0. Therefore, cosA = 1/2, which implies that A = 60°. Since the sum of angles in a triangle is 180°, the other two angles must add up to 120°. This implies that the triangle ABC is an acute-angled triangle. Signup and view all the answers

The sides of a triangle are consecutive integers n, n + 1, and n + 2, and the largest angle is twice the smallest angle. Find the value of n.

Let the smallest angle be x. The largest angle is 2x. The third angle is 180° - x - 2x = 180° - 3x. Using the Law of Cosines, we have: (n + 2)² = n² + (n + 1)² - 2n(n + 1)cos(2x). Similarly, we have: (n + 1)² = n² + (n + 2)² - 2n(n + 2)cos(x). Simplifying these equations and dividing the first equation by the second, we get: [(n + 2)² - n² - (n + 1)²] / [(n + 1)² - n² - (n + 2)²] = cos(2x) / cos(x). Using the double angle formula, cos(2x) = 2cos²(x) - 1, we obtain: (2n + 3) / (2n + 1) = (2cos²(x) - 1) / cos(x). Simplifying and rearranging, we get: 2cos²(x) - (2n + 3)cos(x) + (2n + 1) = 0. Solving this quadratic equation for cos(x), we find cos(x) = 1 or cos(x) = (2n + 1)/2. Since x is an angle in a triangle, 0° < x < 180°. Therefore, cos(x) ≠ 1. Hence, cos(x) = (2n + 1)/2. Using the Law of Cosines again, we have: (n + 1)² = n² + (n + 2)² - 2n(n + 2)[(2n + 1)/2]. Simplifying this equation, we get: 4n² + 12n + 5 = 0. Solving this quadratic equation, we find n = -1/2 or n = -5/2. Since n is a side length of a triangle, it must be positive. Therefore, there is no solution for n. Signup and view all the answers

In a triangle ABC, prove that 1/P1 + 1/P2 + 1/P3 = (a + b + c)/(2ab) * cos²(C/2), where P1, P2, P3 are the altitudes from vertices A, B, and C, respectively, and A denotes the area of the triangle.

We know that A = (1/2) * base * altitude. Therefore, A = (1/2) * BC * P1 = (1/2) * AC * P2 = (1/2) * AB * P3. Solving for P1, P2, and P3, we get: P1 = 2A / BC = 2A / a, P2 = 2A / AC = 2A / b, P3 = 2A / AB = 2A / c. Now, 1/P1 + 1/P2 + 1/P3 = a/(2A) + b/(2A) + c/(2A) = (a + b + c) / (2A). Using the formula for the area of a triangle, A = (1/2) * bc * sinC, we get: 1/P1 + 1/P2 + 1/P3 = (a + b + c) / (bc * sinC). Using the half-angle formula, sinC = 2 * sin(C/2) * cos(C/2), we have: 1/P1 + 1/P2 + 1/P3 = (a + b + c) / (2bc * sin(C/2) * cos(C/2)). Using the formula for the area of a triangle, A = rs, where r is the inradius and s is the semi-perimeter, we get: A = (1/2) * (a + b + c) * r. Substituting this expression into the previous equation, we obtain: 1/P1 + 1/P2 + 1/P3 = (a + b + c) / (2bc * sin(C/2) * cos(C/2)) = (a + b + c) / (2ab * cos²(C/2)). Signup and view all the answers

The triangle ABC satisfies loga² = logb² + logc² - log(2bccosA). What can you say about this triangle?

Using log properties we can simplify the equation to loga² = log(b²c² / 2bccosA) which further reduces to loga² = log(bc / 2cosA). Now we remove the log functions and obtain a² = bc / 2cosA. Using the Law of Cosines, we know that a² = b² + c² - 2bccosA. Substituting this into the previous equation, we get: b² + c² - 2bccosA = bc / 2cosA. Simplifying and multiplying both sides by 2cosA, we get: 2b²cosA + 2c²cosA - 4bccos²A = bc. Rearranging, we get: 2b²cosA + 2c²cosA - bc - 4bccos²A = 0. Factoring by grouping, we obtain: (2cosA - 1)(b² + c² - 2bccosA) = 0. Since b² + c² - 2bccosA = a² > 0, we must have 2cosA - 1 = 0. Therefore, cosA = 1/2, which implies that A = 60°. Since the sum of angles in a triangle is 180°, the other two angles must add up to 120°. This implies that the triangle ABC is an acute-angled triangle. Specifically, it is a 30-60-90 triangle. Signup and view all the answers

In a triangle ABC, if circles with radii r1, r2, and r3 are drawn inside the triangle, each touching two sides and the incircle (with radius r), find the radius of the incircle of triangle ABC.

Let the points of tangency of the incircle with sides BC, AC, and AB be D, E, and F, respectively. Let the points of tangency of the circle with radius r1 with sides BC and AC be G and H, respectively. Similarly, let the points of tangency of the circles with radii r2 and r3 be I, J and K, L, respectively. Since the tangents from a point to a circle are equal, we have: BG = BH = r1, CI = CJ = r2, AK = AL = r3, AD = AE = r + r1, BD = BF = r + r2, CE = CF = r + r3. Now, consider triangle ABC. We have: AB = (r + r2) + (r + r3) = 2r + r2 + r3, BC = (r + r1) + (r + r3) = 2r + r1 + r3, AC = (r + r1) + (r + r2) = 2r + r1 + r2. Adding all these equations, we get: AB + BC + AC = 6r + 2(r1 + r2 + r3) = 2s, where s is the semi-perimeter of triangle ABC. Therefore, r = (s - r1 - r2 - r3) / 3. Hence, the radius of the incircle of triangle ABC is (s - r1 - r2 - r3) / 3. Signup and view all the answers

Study Notes