Q.For any two vectors $\vec{a}$ and $\vec{b}$, we always have $|\vec{a}+\vec{b}|\le|\vec{a}|+|\vec{b}|$ (triangle inequality).

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If the circumcenter of the triangle formed by the points $(1, 2, 3)$, $(3, -1, 5)$ and $(4, 0, -3)$ is $(\alpha, \beta, \gamma)$, then $|\alpha| + |\beta| =$ (A) $3|\gamma|$ (B) $|\gamma|$ (C) $4|\gamma|$ (D) $2|\gamma|$

   ›Reveal solutionSolution

   The circumcenter of a triangle in 3D is the intersection of perpendicular bisectors of two sides; solving the system gives $(\alpha,\beta,\gamma) = (3, 1, 1)$, so $|\alpha|+|\beta| = 4 = 4|\gamma|$, matching option (C).

   Concept & Intuition

   In 3D, the circumcenter is the point equidistant from all three vertices. Instead of solving three distance equations directly (which leads to messy squares), we use the fact that the circumcenter lies on the perpendicular bisector of each side. For a side $AB$, the perpendicular bisector is the set of points $X$ such that $(X - M_{AB}) \cdot (B - A) = 0$, where $M_{AB}$ is the midpoint. Intersecting two such planes gives a line; intersecting with a third plane (or using the distance equality) pins down the unique point.

   ---

   Step-by-step solution

   1. Label the points

      Let $A = (1,2,3)$, $B = (3,-1,5)$, $C = (4,0,-3)$.

      The circumcenter $O = (\alpha,\beta,\gamma)$ satisfies $OA = OB = OC$.

   2. Use two perpendicular bisector conditions

      For side $AB$:

      Midpoint $M_{AB} = \left(\frac{1+3}{2}, \frac{2+(-1)}{2}, \frac{3+5}{2}\right) = (2, 0.5, 4)$.

      Vector $\overrightarrow{AB} = (3-1, -1-2, 5-3) = (2, -3, 2)$.

      Condition: $(O - M_{AB}) \cdot \overrightarrow{AB} = 0$

      $\Rightarrow (\alpha-2, \beta-0.5, \gamma-4) \cdot (2, -3, 2) = 0$

      $\Rightarrow 2(\alpha-2) -3(\beta-0.5) + 2(\gamma-4) = 0$

      $\Rightarrow 2\alpha -4 -3\beta +1.5 + 2\gamma -8 = 0$

      $\Rightarrow 2\alpha -3\beta + 2\gamma = 10.5$

      Multiply by 2 to avoid decimals:

   $4\alpha -6\beta + 4\gamma = 21 \quad \text{(1)}$

   3. Second side: $AC$ Midpoint $M_{AC} = \left(\frac{1+4}{2}, \frac{2+0}{2}, \frac{3+(-3)}{2}\right) = (2.5, 1, 0)$. $\overrightarrow{AC} = (4-1, 0-2, -3-3) = (3, -2, -6)$. Condition: $(O - M_{AC}) \cdot \overrightarrow{AC} = 0$ $\Rightarrow (\alpha-2.5, \beta-1, \gamma-0) \cdot (3, -2, -6) = 0$ $\Rightarrow 3(\alpha-2.5) -2(\beta-1) -6\gamma = 0$ $\Rightarrow 3\alpha -7.5 -2\beta +2 -6\gamma = 0$ $\Rightarrow 3\alpha -2\beta -6\gamma = 5.5$ Multiply by 2:

   $6\alpha -4\beta -12\gamma = 11 \quad \text{(2)}$

   4. Third condition: $OA = OB$ (distance equality)

      $OA^2 = OB^2$ gives:

      $(\alpha-1)^2 + (\beta-2)^2 + (\gamma-3)^2 = (\alpha-3)^2 + (\beta+1)^2 + (\gamma-5)^2$

      Expand:

      Left: $\alpha^2 -2\alpha +1 + \beta^2 -4\beta +4 + \gamma^2 -6\gamma +9$

      Right: $\alpha^2 -6\alpha +9 + \beta^2 +2\beta +1 + \gamma^2 -10\gamma +25$

      Cancel $\alpha^2,\beta^2,\gamma^2$:

      $-2\alpha -4\beta -6\gamma +14 = -6\alpha +2\beta -10\gamma +35$

      Bring terms: $(-2\alpha+6\alpha) + (-4\beta-2\beta) + (-6\gamma+10\gamma) = 35-14$

      $\Rightarrow 4\alpha -6\beta +4\gamma = 21$

      This is exactly equation (1)! So it’s not independent — we need a third distinct condition.

   5. Use $OA = OC$ instead

      $(\alpha-1)^2 + (\beta-2)^2 + (\gamma-3)^2 = (\alpha-4)^2 + (\beta-0)^2 + (\gamma+3)^2$

      Expand:

      Left: same as before: $\alpha^2 -2\alpha +1 + \beta^2 -4\beta +4 + \gamma^2 -6\gamma +9$

      Right: $\alpha^2 -8\alpha +16 + \beta^2 + \gamma^2 +6\gamma +9$

      Cancel: $-2\alpha -4\beta -6\gamma +14 = -8\alpha +6\gamma +25$

      $\Rightarrow (-2\alpha+8\alpha) -4\beta + (-6\gamma-6\gamma) = 25-14$

      $\Rightarrow 6\alpha -4\beta -12\gamma = 11$

      This is exactly equation (2). So again not independent — the two perpendicular bisector planes already encode both distance equalities.

   6. We need a third plane: use side $BC$

      Midpoint $M_{BC} = \left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5+(-3)}{2}\right) = (3.5, -0.5, 1)$.

      $\overrightarrow{BC} = (4-3, 0-(-1), -3-5) = (1, 1, -8)$.

      Condition: $(O - M_{BC}) \cdot \overrightarrow{BC} = 0$

      $\Rightarrow (\alpha-3.5, \beta+0.5, \gamma-1) \cdot (1, 1, -8) = 0$

      $\Rightarrow (\alpha-3.5) + (\beta+0.5) -8(\gamma-1) = 0$

      $\Rightarrow \alpha + \beta -8\gamma -3.5 +0.5 +8 = 0$

      $\Rightarrow \alpha + \beta -8\gamma +5 = 0$

      So

   $\alpha + \beta -8\gamma = -5 \quad \text{(3)}$

   7. Solve the system

      From (1): $4\alpha -6\beta +4\gamma = 21$

      From (2): $6\alpha -4\beta -12\gamma = 11$

      From (3): $\alpha + \beta -8\gamma = -5$

      Solve (3) for $\alpha$: $\alpha = -5 - \beta + 8\gamma$.

      Substitute into (1):

      $4(-5 - \beta + 8\gamma) -6\beta +4\gamma = 21$

      $\Rightarrow -20 -4\beta +32\gamma -6\beta +4\gamma = 21$

      $\Rightarrow -10\beta +36\gamma = 41$

      $\Rightarrow -10\beta = 41 - 36\gamma$

      $\Rightarrow \beta = \frac{36\gamma - 41}{10}$

      Substitute into (2):

      $6(-5 - \beta + 8\gamma) -4\beta -12\gamma = 11$

      $\Rightarrow -30 -6\beta +48\gamma -4\beta -12\gamma = 11$

      $\Rightarrow -10\beta +36\gamma = 41$ (same as above, consistent).

      Now use (3) again with $\beta$ expression:

      $\alpha = -5 - \frac{36\gamma - 41}{10} + 8\gamma = -5 - \frac{36\gamma}{10} + \frac{41}{10} + 8\gamma$

      $= -5 + 4.1 + \left(8 - 3.6\right)\gamma = -0.9 + 4.4\gamma$

      So $\alpha = \frac{44\gamma - 9}{10}$.

      Now use distance equality $OA = OC$ (or any vertex) to find $\gamma$. Use $OA^2 = OB^2$ already used, but we haven't used a numeric distance — pick $OA^2 = OB^2$ gave identity. Use $OA^2 = OC^2$ also identity. So we need one more: use $OB^2 = OC^2$? That will also be dependent. Actually we have three planes; they should intersect at a point. Solve (1) and (3) for $\alpha,\beta$ in terms of $\gamma$, then plug into (2) to find $\gamma$.

      From (3): $\alpha = -5 - \beta + 8\gamma$.

      Substitute into (1): $4(-5 - \beta + 8\gamma) -6\beta +4\gamma = 21$

      $\Rightarrow -20 -4\beta +32\gamma -6\beta +4\gamma = 21$

      $\Rightarrow -10\beta +36\gamma = 41$

      $\Rightarrow \beta = \frac{36\gamma - 41}{10}$.

      Substitute $\beta$ into (2): $6\alpha -4\left(\frac{36\gamma - 41}{10}\right) -12\gamma = 11$

      But $\alpha = -5 - \frac{36\gamma - 41}{10} + 8\gamma = \frac{-50 -36\gamma +41 +80\gamma}{10} = \frac{-9 +44\gamma}{10}$.

      Then (2): $6\cdot\frac{44\gamma -9}{10} - \frac{4(36\gamma -41)}{10} -12\gamma = 11$

      Multiply by 10: $6(44\gamma -9) -4(36\gamma -41) -120\gamma = 110$

      $\Rightarrow 264\gamma -54 -144\gamma +164 -120\gamma = 110$

      $\Rightarrow (264-144-120)\gamma + ( -54+164) = 110$ …

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2. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.In a $\triangle ABC$, if $\cos A + \cos B + \cos C = a + b \sin \dfrac{A}{2} - \sin \dfrac{B}{2} - \sin \dfrac{C}{2}$, then $a + b =$ (A) 3 (B) 0 (C) 1 (D) 5

   ›Reveal solutionSolution

   cos A + cos B + cos C = 1 + 4 sin(A/2)sin(B/2)sin(C/2) ⇒ a=1, b=4, a+b=5.

   For any triangle, cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2). Comparing the right-hand side of the given relation, the constant term identifies a = 1 and the coefficient of the half-angle sine pr …

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3. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.In any $\triangle ABC$, $\dfrac{1 + \cos(A-B)\cdot\cos C}{1 + \cos(A-C)\cdot\cos B} =$ (A) $\dfrac{a^2 + c^2}{b^2 + c^2}$ (B) $\dfrac{b^2 + c^2}{b^2 + a^2}$ (C) $\dfrac{a^2 + c^2}{a^2 + b^2}$ (D) $\dfrac{a^2 + b^2}{a^2 + c^2}$

   ›Reveal solutionSolution

   By using the angle sum property of a triangle ($A+B+C=\pi$) to simplify the cosine terms and then applying the identity $\cos(X-Y)\cos(X+Y) = \cos^2 X - \sin^2 Y$, the expression simplifies to $\dfrac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C}$. Finally, using the Sine Rule, this converts to $\boxed{\dfrac{a^2 + b^2}{a^2 + c^2}}$.

   In any triangle $ABC$, the sum of the angles is always $\pi$ radians (or $180^\circ$). This fundamental property, $A+B+C = \pi$, allows us to relate the angles to each other. For instance, $C = \pi - (A+B)$, which means $\cos C = \cos(\pi - (A+B)) = -\cos(A+B)$. Similarly, $\cos B = -\cos(A+C)$. This substitution is the key to simplifying the given expression.

   Once the expression is simplified in terms of sines of angles, we can use the Sine Rule, which states that the ratio of a side length to the sine of its opposite angle is constant for all sides of a triangle: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k$ (where $k$ is a constant). This allows us to convert expressions involving $\sin A$, $\sin B$, $\sin C$ into expressions involving the side lengths $a$, $b$, $c$.

   Let's break down the simplification:

   1. Simplify the numerator:

      The numerator is $1 + \cos(A-B)\cdot\cos C$.

      Since $A+B+C = \pi$, we have $C = \pi - (A+B)$.

      Therefore, $\cos C = \cos(\pi - (A+B)) = -\cos(A+B)$.

      Substitute this into the numerator:

      $1 + \cos(A-B)\cdot(-\cos(A+B)) = 1 - \cos(A-B)\cos(A+B)$.

      A useful trigonometric identity is $\cos(X-Y)\cos(X+Y) = \cos^2 X - \sin^2 Y$.

      Applying this identity with $X=A$ and $Y=B$:

      $1 - (\cos^2 A - \sin^2 B)$

      $= 1 - \cos^2 A + \sin^2 B$

      Since $1 - \cos^2 A = \sin^2 A$, the numerator simplifies to:

      $\sin^2 A + \sin^2 B$.

   2. Simplify the denominator:

      The denominator is $1 + \cos(A-C)\cdot\cos B$.

      Similarly, since $A+B+C = \pi$, we have $B = \pi - (A+C)$.

      Therefore, $\cos B = \cos(\pi - (A+C)) = -\cos(A+C)$.

      Substitute this into the denominator:

      $1 + \cos(A-C)\cdot(-\cos(A+C)) = 1 - \cos(A-C)\cos(A+C)$.

      Applying the same identity $\cos(X-Y)\cos(X+Y) = \cos^2 X - \sin^2 Y$ with $X=A$ and $Y=C$:

      $1 - (\cos^2 A - \sin^2 C)$

      $= 1 - \cos^2 A + \sin^2 C$

      Since $1 - \cos^2 A = \sin^2 A$, the denominator simplifies to:

      $\sin^2 A + \sin^2 C$.

   3. Form the ratio: …

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4. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.In a triangle, if the length of the sides $a, b$ and $c$ are three consecutive natural numbers and $a < b < c$, then $(\cos A + \cos B + \cos C)2abc =$ (A) $3b(b^2 - 2)$ (B) $3b^3 + 6b^2 + 3b$ (C) $(3b + 2)(3b - 2)b$ (D) $(b - 1)b(b + 1)$

   ›Reveal solutionSolution

   The key idea is to express the cosines using the law of cosines and the sides as consecutive natural numbers $a = b-1$, $b$, $c = b+1$. After simplification, the expression $(\cos A + \cos B + \cos C)2abc$ reduces to $3b(b^2 - 2)$, which matches option (A).

   The problem gives a triangle with sides $a, b, c$ as three consecutive natural numbers, with $a < b < c$. So we can set $a = b-1$, $b = b$, and $c = b+1$, where $b$ is a natural number greater than 1 (to ensure $a > 0$). The expression to evaluate is $(\cos A + \cos B + \cos C) \cdot 2abc$.

   The natural instinct is to compute each cosine using the law of cosines, then sum them, and multiply by $2abc$. The law of cosines relates each angle to the three sides, so everything becomes algebraic in terms of $b$. The beauty is that the factor $2abc$ will cancel denominators neatly, leaving a polynomial in $b$.

   Let’s proceed step by step.

   1. Set up the sides. Since $a, b, c$ are consecutive natural numbers and $a < b < c$, we have:

   $a = b-1, \quad b = b, \quad c = b+1.$

   The triangle inequality must hold: $a + b > c$ gives $(b-1) + b > b+1 \Rightarrow 2b-1 > b+1 \Rightarrow b > 2$, so $b \ge 3$ (natural numbers). This is fine.

   2. Write the law of cosines for each angle. For angle $A$ (opposite side $a$):

   $\cos A = \frac{b^2 + c^2 - a^2}{2bc}.$

   For angle $B$ (opposite side $b$):

   $\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$

   For angle $C$ (opposite side $c$):

   $\cos C = \frac{a^2 + b^2 - c^2}{2ab}.$

   3. Substitute the side values.

      Compute each numerator:

      • $b^2 + c^2 - a^2 = b^2 + (b+1)^2 - (b-1)^2$ Expand: $(b+1)^2 = b^2 + 2b + 1$, $(b-1)^2 = b^2 - 2b + 1$. So numerator = $b^2 + (b^2 + 2b + 1) - (b^2 - 2b + 1) = b^2 + b^2 + 2b + 1 - b^2 + 2b - 1 = b^2 + 4b$.
      • $a^2 + c^2 - b^2 = (b-1)^2 + (b+1)^2 - b^2$ = $(b^2 - 2b + 1) + (b^2 + 2b + 1) - b^2 = b^2 + 2$.
      • $a^2 + b^2 - c^2 = (b-1)^2 + b^2 - (b+1)^2$ = $(b^2 - 2b + 1) + b^2 - (b^2 + 2b + 1) = b^2 - 4b$.

      The denominators:

      • For $\cos A$: $2bc = 2b(b+1)$.
      • For $\cos B$: $2ac = 2(b-1)(b+1) = 2(b^2 - 1)$.
      • For $\cos C$: $2ab = 2(b-1)b$.

      So:

   $\cos A = \frac{b^2 + 4b}{2b(b+1)} = \frac{b(b+4)}{2b(b+1)} = \frac{b+4}{2(b+1)}.$

   $\cos B = \frac{b^2 + 2}{2(b^2 - 1)}.$

   $\cos C = \frac{b^2 - 4b}{2b(b-1)} = \frac{b(b-4)}{2b(b-1)} = \frac{b-4}{2(b-1)}.$

   4. Sum the cosines.

   $\cos A + \cos B + \cos C = \frac{b+4}{2(b+1)} + \frac{b^2 + 2}{2(b^2 - 1)} + \frac{b-4}{2(b-1)}.$

   Factor out $\frac{1}{2}$:

   $= \frac{1}{2} \left[ \frac{b+4}{b+1} + \frac{b^2 + 2}{b^2 - 1} + \frac{b-4}{b-1} \right].$ …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $z$ is a complex number such that $\left| 9z + \frac{1}{z} \right| = 8$, then the maximum value of $|z|$ is (A) $\frac{1}{8}$ (B) $\frac{2}{5}$ (C) $\frac{7}{2}$ (D) $1$

   ›Reveal solutionSolution

   Using the reverse triangle inequality on $\left|9z + \frac{1}{z}\right| = 8$ gives a pair of quadratic inequalities in $|z|$, whose combined solution is $\frac{1}{9} \le |z| \le 1$. The maximum value of $|z|$ is $1$, corresponding to option (D).

   ---

   Concept and Intuition

   We are given a constraint involving a complex number $z$ and its reciprocal: $\left| 9z + \frac{1}{z} \right| = 8$. We want the largest possible magnitude $|z|$. The reverse triangle inequality — for any complex numbers $a$ and $b$, $|a+b| \ge \big||a| - |b|\big|$ — lets us turn the given equality into an inequality involving only $|z|$, which we can solve.

   ---

   Step-by-step solution

   1. Set up notation.

      Let $r = |z| > 0$ (since $z$ appears in the denominator, $z \neq 0$). Then $|9z| = 9r$ and $\left|\frac{1}{z}\right| = \frac{1}{r}$.

   2. Apply the reverse triangle inequality.

   $\left| 9z + \frac{1}{z} \right| \ge \left| 9r - \frac{1}{r} \right|.$

   Since the left side equals $8$:

   $\left| 9r - \frac{1}{r} \right| \le 8.$

   3. Remove the absolute value.

   $-8 \le 9r - \frac{1}{r} \le 8.$

   4. Solve the right-hand inequality.

   $9r - \frac{1}{r} \le 8 \quad \Rightarrow \quad 9r^2 - 8r - 1 \le 0.$

   The roots of $9r^2 - 8r - 1 = 0$ are

   $r = \frac{8 \pm \sqrt{64 + 36}}{18} = \frac{8 \pm 10}{18} \Rightarrow r = 1 \text{ or } r = -\frac{1}{9}.$

   Since the quadratic opens upward, $9r^2-8r-1\le0$ holds between the roots: $-\frac{1}{9} \le r \le 1$. With $r>0$, this gives $0 < r \le 1$.

   5. Solve the left-hand inequality.

   $9r - \frac{1}{r} \ge -8 \quad \Rightarrow \quad 9r^2 + 8r - 1 \ge 0.$

   The roots of $9r^2 + 8r - 1 = 0$ are

   $r = \frac{-8 \pm \sqrt{64 + 36}}{18} = \frac{-8 \pm 10}{18} \Rightarrow r = \frac{1}{9} \text{ or } r = -1.$

   Since the quadratic opens upward, $9r^2+8r-1\ge0$ holds outside the roots: $r \le -1$ or $r \ge \frac{1}{9}$. With $r>0$, this gives $r \ge \frac{1}{9}$.

   6. Combine the constraints.

   $\frac{1}{9} \le r \le 1.$

   So the maximum possible $|z|$ is $1$. …

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.A$(2,3,k)$, B$(-1,k,-1)$, C$(4,-3,2)$ are the vertices of $\triangle ABC$. If $AB = AC$ and $k > 0$, then $ABC$ is (A) an equilateral triangle (B) a right-angled isosceles triangle (C) an isosceles triangle but not right angled (D) an obtuse angled isosceles triangle

   ›Reveal solutionSolution

   The key is to use the distance formula to enforce $AB = AC$, solve for $k$, then check side lengths and the Pythagorean relation to classify the triangle. The result is a right-angled isosceles triangle, so option (B) is correct.

   We are given three points in 3D space:

   $A(2,3,k)$, $B(-1,k,-1)$, $C(4,-3,2)$.

   The condition $AB = AC$ with $k > 0$ lets us find $k$. Then we can compute all three side lengths and decide the triangle type.

   ---

   1. Write the distance formulas for $AB$ and $AC$.

   The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is

   $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.$

   So

   $AB^2 = (-1-2)^2 + (k-3)^2 + (-1-k)^2 = 9 + (k-3)^2 + (k+1)^2,$

   $AC^2 = (4-2)^2 + (-3-3)^2 + (2-k)^2 = 4 + 36 + (k-2)^2.$

   2. Set $AB^2 = AC^2$ and solve for $k$.

   $9 + (k-3)^2 + (k+1)^2 = 40 + (k-2)^2.$

   Expand the squares:

   $(k-3)^2 = k^2 - 6k + 9$,

   $(k+1)^2 = k^2 + 2k + 1$,

   $(k-2)^2 = k^2 - 4k + 4$.

   Left side: $9 + (k^2 - 6k + 9) + (k^2 + 2k + 1) = 2k^2 - 4k + 19$.

   Right side: $40 + (k^2 - 4k + 4) = k^2 - 4k + 44$.

   Equate:

   $2k^2 - 4k + 19 = k^2 - 4k + 44.$

   Cancel $-4k$ on both sides:

   $2k^2 + 19 = k^2 + 44 \quad\Rightarrow\quad k^2 = 25.$

   Since $k > 0$, we get $k = 5$.

   Tip

   Notice the $-4k$ terms cancel immediately — this is a clean quadratic, not a linear trap.

   3. Compute the side lengths with $k=5$.

   $A(2,3,5)$, $B(-1,5,-1)$, $C(4,-3,2)$.

   $AB^2 = 9 + (5-3)^2 + (-1-5)^2 = 9 + 4 + 36 = 49$, so $AB = 7$. …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.In a quadrilateral $ABCD$, $|A| = \dfrac{2\pi}{3}$ and $AC$ is the bisector of angle $|A|$. If $15|AC| = 5|AD| = 3|AB|$, then angle between $\overline{AB}$ and $\overline{BC}$ is (A) $\cos^{-1}\left(\dfrac{\sqrt{3}}{\sqrt{7}}\right)$ (B) $\cos^{-1}\left(\dfrac{3\sqrt{3}}{2\sqrt{7}}\right)$ (C) $\cos^{-1}\left(\dfrac{4\sqrt{3}}{5\sqrt{7}}\right)$ (D) $\cos^{-1}\left(\dfrac{3\sqrt{3}}{4\sqrt{7}}\right)$

   ›Reveal solutionSolution

   Using the angle bisector theorem and the law of cosines in triangles formed by the diagonal, we find the angle between AB and BC is $\cos^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right)$, which corresponds to option (B).

   We are given quadrilateral $ABCD$ with $\angle A = \frac{2\pi}{3}$ and $AC$ bisecting this angle. That means $\angle BAC = \angle CAD = \frac{\pi}{3}$. Also, $15|AC| = 5|AD| = 3|AB|$. Let’s set a common variable to simplify.

   Let $|AC| = k$. Then:

   • $15k = 5|AD| \implies |AD| = 3k$
   • $15k = 3|AB| \implies |AB| = 5k$

   So we have triangle $ABC$ with $AB = 5k$, $AC = k$, and $\angle BAC = 60^\circ$. Similarly, triangle $ACD$ has $AD = 3k$, $AC = k$, and $\angle CAD = 60^\circ$.

   We need the angle between $\overline{AB}$ and $\overline{BC}$, which is $\angle ABC$ (since $B$ is the vertex where $AB$ and $BC$ meet). We can find $BC$ using the law of cosines in triangle $ABC$, then use the law of cosines again to find $\angle ABC$.

   1. Find $BC$ in triangle $ABC$ By law of cosines:

   $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$

   Substitute $AB = 5k$, $AC = k$, $\cos 60^\circ = \frac12$:

   $BC^2 = (5k)^2 + k^2 - 2(5k)(k)\left(\frac12\right) = 25k^2 + k^2 - 5k^2 = 21k^2$

   So $BC = \sqrt{21}\,k$.

   2. Find $\angle ABC$ in triangle $ABC$ Use law of cosines again, solving for $\cos(\angle ABC)$:

   $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$

   Substitute $AC = k$, $AB = 5k$, $BC = \sqrt{21}\,k$:

   $k^2 = (5k)^2 + (\sqrt{21}k)^2 - 2(5k)(\sqrt{21}k)\cos(\angle ABC)$

   $k^2 = 25k^2 + 21k^2 - 10\sqrt{21}\,k^2 \cos(\angle ABC)$

   $k^2 = 46k^2 - 10\sqrt{21}\,k^2 \cos(\angle ABC)$

   Divide through by $k^2$ (nonzero):

   $1 = 46 - 10\sqrt{21} \cos(\angle ABC)$

   $10\sqrt{21} \cos(\angle ABC) = 45$

   $$$$

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.In a quadrilateral ABCD, $\angle A = \dfrac{2\pi}{3}$ and AC is the bisector of angle $\angle A$. If $15|AC| = 5|AD| = 3|AB|$, then angle between $\overline{AB}$ and $\overline{BC}$ is (A) $\cos^{-1}\left(\dfrac{\sqrt{3}}{\sqrt{7}}\right)$ (B) $\cos^{-1}\left(\dfrac{3\sqrt{3}}{4\sqrt{7}}\right)$ (C) $\cos^{-1}\left(\dfrac{3\sqrt{3}}{2\sqrt{7}}\right)$ (D) $\cos^{-1}\left(\dfrac{4\sqrt{3}}{5\sqrt{7}}\right)$

   ›Reveal solutionSolution

   With $\angle BAC=60^\circ$ and the given side ratios, the law of cosines in $\triangle ABC$ gives the angle at $B$ as $\cos^{-1}\!\left(\dfrac{3\sqrt3}{2\sqrt7}\right)$ — option (C).

   Concept

   The required angle between $\overline{AB}$ and $\overline{BC}$ is angle $B$ of triangle $ABC$. Since $AC$ bisects $\angle A=\tfrac{2\pi}{3}=120^\circ$, the angle $\angle BAC=60^\circ$. Knowing $AB$ and $AC$, apply the law of cosines twice.

   Step-by-step solution

   1. Assign lengths. Let $15|AC|=5|AD|=3|AB|=k$, so

   $|AB|=\frac{k}{3},\qquad |AC|=\frac{k}{15}.$

   2. Find $BC$ in $\triangle ABC$ with $\angle BAC=60^\circ$:

   $BC^2=AB^2+AC^2-2\,AB\cdot AC\cos60^\circ=\frac{k^2}{9}+\frac{k^2}{225}-\frac{k^2}{45}.$

   Over denominator $225$: $\dfrac{25+1-5}{225}k^2=\dfrac{21k^2}{225}=\dfrac{7k^2}{75}$, so $BC=\dfrac{k\sqrt7}{5\sqrt3}$.

   3. Find $\cos B$ by the law of cosines: $\cos B=\frac{AB^2+BC^2-AC^2}{2\,AB\cdot BC}.$ …

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9. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $z_1 = 1 - 2i$, $z_2 = 1 + i$ and $z_3 = 3 + 4i$, then $\left| \left( \frac{1}{z_1} + \frac{2}{z_2} \right) \frac{z_3}{z_2} \right| =$ (A) $\frac{\sqrt{7}}{2}$ (B) $\frac{\sqrt{5}}{2}$ (C) $\sqrt{\frac{45}{2}}$ (D) $\frac{\sqrt{15}}{2}$

   ›Reveal solutionSolution

   Simplifying gives $\dfrac{9-3i}{2}$, whose modulus is $\dfrac{\sqrt{90}}{2} = \sqrt{\dfrac{45}{2}}$ — option (C).

   With $z_1 = 1-2i$, $z_2 = 1+i$, $z_3 = 3+4i$, first compute the bracket.

   $\frac{1}{z_1} = \frac{1}{1-2i} = \frac{1+2i}{(1)^2+(2)^2} = \frac{1+2i}{5},$

   $\frac{2}{z_2} = \frac{2}{1+i} = \frac{2(1-i)}{(1)^2+(1)^2} = 1-i = \frac{5-5i}{5}.$

   Add them:

   $\frac{1}{z_1}+\frac{2}{z_2} = \frac{(1+2i)+(5-5i)}{5} = \frac{6-3i}{5}.$

   Next,

   $\frac{z_3}{z_2} = \frac{3+4i}{1+i} = \frac{(3+4i)(1-i)}{2} = \frac{3-3i+4i-4i^2}{2} = \frac{7+i}{2}.$

   Multiply: …

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10. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
    Q.In triangle ABC, if $A = \dfrac{\pi}{3}$ and $B = \dfrac{\pi}{4}$ then $\dfrac{a^2 - b^2}{c^2} =$ (A) $2 - \sqrt{3}$ (B) $2 + \sqrt{3}$ (C) $\sqrt{2} - 1$ (D) $\sqrt{2} + 1$

    ›Reveal solutionSolution

    Use the Law of Sines to express sides in terms of sines of angles, then simplify the given expression using known sine values. The result is $\frac{2 - \sqrt{3}}{2}$, which matches option (A).

    The key here is that in any triangle, sides are proportional to the sines of their opposite angles. That’s the Law of Sines:

    $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

    where $R$ is the circumradius. So instead of working with side lengths directly, we can replace $a$, $b$, $c$ with $\sin A$, $\sin B$, $\sin C$ — the factor $2R$ cancels out in a ratio like this.

    We are given $A = \frac{\pi}{3}$ and $B = \frac{\pi}{4}$. The third angle $C$ follows from $A + B + C = \pi$:

    $C = \pi - \frac{\pi}{3} - \frac{\pi}{4} = \frac{12\pi - 4\pi - 3\pi}{12} = \frac{5\pi}{12}$

    Now let’s work through the expression step by step.

    1. Express the ratio in terms of sines Using the Law of Sines: $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$. Then

    $\frac{a^2 - b^2}{c^2} = \frac{(2R)^2 (\sin^2 A - \sin^2 B)}{(2R)^2 \sin^2 C} = \frac{\sin^2 A - \sin^2 B}{\sin^2 C}$

    2. Simplify the numerator using a trigonometric identity Recall: $\sin^2 X - \sin^2 Y = \sin(X+Y) \sin(X-Y)$. So

    $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$

    Here $A+B = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$ and $A-B = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$.

    3. Plug in the known sine values

       We have:

       • $\sin C = \sin \frac{5\pi}{12}$
       • $\sin(A+B) = \sin \frac{7\pi}{12}$
       • $\sin(A-B) = \sin \frac{\pi}{12}$

       Notice that $\frac{7\pi}{12} = \pi - \frac{5\pi}{12}$, so $\sin \frac{7\pi}{12} = \sin \frac{5\pi}{12}$.

       Therefore the expression becomes:

    $\frac{\sin \frac{5\pi}{12} \cdot \sin \frac{\pi}{12}}{\sin^2 \frac{5\pi}{12}} = \frac{\sin \frac{\pi}{12}}{\sin \frac{5\pi}{12}}$

    4. Evaluate the sines Use known exact values: …

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11. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.In $\triangle ABC$, if $B + C = 72^\circ$, then $\left(1 + \frac{a}{c} + \frac{b}{c}\right)\left(1 + \frac{c}{b} - \frac{a}{b}\right) =$ (A) $2 + \sqrt{5}$ (B) $\frac{\sqrt{5} + 1}{2\sqrt{2}}$ (C) $\frac{\sqrt{5} - 2}{4}$ (D) $\frac{5 - \sqrt{5}}{2}$

    ›Reveal solutionSolution

    Use the Law of Sines to rewrite the expression in terms of sines of angles, then apply the given angle sum $B+C=72^\circ$ and known trigonometric identities to simplify to a constant. The final value is $\frac{5-\sqrt{5}}{2}$.

    The key insight is that the expression involves ratios of sides $a, b, c$ of a triangle. In any triangle, side lengths are proportional to the sines of the opposite angles via the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ (where $R$ is the circumradius). This lets us replace every side ratio with a ratio of sines, turning the algebraic expression into a purely trigonometric one. Then the condition $B+C=72^\circ$ gives $A = 180^\circ - (B+C) = 108^\circ$, so we know all angles in terms of $B$ and $C$ — but we won't need their individual values because the expression simplifies using sum-to-product identities.

    1. Rewrite using the Law of Sines. Let $a = k\sin A$, $b = k\sin B$, $c = k\sin C$ where $k=2R$. Then:

    $\frac{a}{c} = \frac{\sin A}{\sin C},\quad \frac{b}{c} = \frac{\sin B}{\sin C},\quad \frac{c}{b} = \frac{\sin C}{\sin B},\quad \frac{a}{b} = \frac{\sin A}{\sin B}.$

    The given expression becomes:

    $\left(1 + \frac{\sin A}{\sin C} + \frac{\sin B}{\sin C}\right)\left(1 + \frac{\sin C}{\sin B} - \frac{\sin A}{\sin B}\right).$

    2. Combine terms over common denominators. First factor:

    $1 + \frac{\sin A}{\sin C} + \frac{\sin B}{\sin C} = \frac{\sin C + \sin A + \sin B}{\sin C}.$

    Second factor:

    $1 + \frac{\sin C}{\sin B} - \frac{\sin A}{\sin B} = \frac{\sin B + \sin C - \sin A}{\sin B}.$

    So the product is:

    $\frac{(\sin A + \sin B + \sin C)(\sin B + \sin C - \sin A)}{\sin B \sin C}.$

    3. Use the angle sum $A+B+C=180^\circ$ and $B+C=72^\circ$. Hence $A = 108^\circ$. We now simplify the numerator using trigonometric identities. Notice that $\sin B + \sin C - \sin A$ can be rewritten. Since $A = 180^\circ - (B+C)$, we have $\sin A = \sin(B+C)$. So:

    $\sin B + \sin C - \sin A = \sin B + \sin C - \sin(B+C).$

    Also, $\sin A + \sin B + \sin C = \sin(B+C) + \sin B + \sin C$.

    The product becomes:

    $\frac{[\sin(B+C) + \sin B + \sin C]\,[\sin B + \sin C - \sin(B+C)]}{\sin B \sin C}.$

    This is of the form $(X+Y)(X-Y)$ where $X = \sin B + \sin C$ and $Y = \sin(B+C)$. So:

    $(X+Y)(X-Y) = X^2 - Y^2 = (\sin B + \sin C)^2 - \sin^2(B+C).$

    4. Apply sum-to-product identities. $\sin B + \sin C = 2\sin\frac{B+C}{2}\cos\frac{B-C}{2}$. Since $B+C=72^\circ$, $\frac{B+C}{2}=36^\circ$. So:

    $\sin B + \sin C = 2\sin 36^\circ \cos\frac{B-C}{2}.$

    Also $\sin(B+C) = \sin 72^\circ$.

    The numerator is:

    $4\sin^2 36^\circ \cos^2\frac{B-C}{2} - \sin^2 72^\circ.$

    5. Eliminate the dependence on $B-C$. The denominator is $\sin B \sin C$. Using the product-to-sum identity:

    $\sin B \sin C = \frac{1}{2}[\cos(B-C) - \cos(B+C)] = \frac{1}{2}[\cos(B-C) - \cos 72^\circ].$

    And $\cos^2\frac{B-C}{2} = \frac{1+\cos(B-C)}{2}$. So the numerator becomes:

    $4\sin^2 36^\circ \cdot \frac{1+\cos(B-C)}{2} - \sin^2 72^\circ = 2\sin^2 36^\circ (1+\cos(B-C)) - \sin^2 72^\circ.$

    The whole expression is:

    $\frac{2\sin^2 36^\circ (1+\cos(B-C)) - \sin^2 72^\circ}{\frac{1}{2}[\cos(B-C) - \cos 72^\circ]}.$

    6. Use known exact values.

       $\sin 36^\circ = \frac{\sqrt{10-2\sqrt{5}}}{4}$, so $\sin^2 36^\circ = \frac{10-2\sqrt{5}}{16} = \frac{5-\sqrt{5}}{8}$.

       $\sin 72^\circ = \cos 18^\circ = \frac{\sqrt{10+2\sqrt{5}}}{4}$, so $\sin^2 72^\circ = \frac{10+2\sqrt{5}}{16} = \frac{5+\sqrt{5}}{8}$.

       $\cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5}-1}{4}$.

       Substitute:

    $\text{Numerator} = 2\cdot\frac{5-\sqrt{5}}{8}(1+\cos(B-C)) - \frac{5+\sqrt{5}}{8} = \frac{5-\sqrt{5}}{4}(1+\cos(B-C)) - \frac{5+\sqrt{5}}{8}.$

    Denominator = $\frac{1}{2}\left(\cos(B-C) - \frac{\sqrt{5}-1}{4}\right) = \frac{1}{2}\cos(B-C) - \frac{\sqrt{5}-1}{8}$.

    7. Simplify algebraically. Write numerator over common denominator 8: $$\frac{2(5-\sqrt{5})(1+\cos(B-C)) - (5+\sqrt{5})}{8} = \frac{(10-2\sqrt{5})(1+\cos(B-C)) - 5 - \sqrt{5}}{8}. …$$

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12. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.In a $\triangle ABC$, the sides b, c are fixed. In measuring angle A, if there is an error of $\delta A$, then the percentage error in measuring the length of the side a is (A) $\dfrac{2\Delta \delta A}{R \sin A} \times 100$ (B) $2 \times \dfrac{\delta A}{A} \times 100$ (C) $\dfrac{\Delta \delta A}{2R^2 \sin^2 A} \times 100$ (D) $\dfrac{\Delta^2 \delta A}{R \sin A} \times 100$

    ›Reveal solutionSolution

    Differentiating $a^2 = b^2 + c^2 - 2bc\cos A$ gives $\dfrac{da}{a} = \dfrac{\Delta\,\delta A}{2R^2\sin^2 A}$, so the percentage error is $\dfrac{\Delta\,\delta A}{2R^2\sin^2 A}\times 100$.

    Relating $a$ to the fixed sides and angle $A$.

    By the cosine rule, with $b,c$ fixed,

    $a^2 = b^2 + c^2 - 2bc\cos A.$

    Differentiating (only $a$ and $A$ vary):

    $2a\,da = 2bc\sin A\,\delta A \quad\Rightarrow\quad da = \frac{bc\sin A}{a}\,\delta A.$

    Percentage error in $a$.

    $\frac{da}{a} = \frac{bc\sin A}{a^2}\,\delta A.$

    Use the area relation $\Delta = \tfrac{1}{2}bc\sin A$, so $bc\sin A = 2\Delta$, and the law of sines $a = 2R\sin A$, so $a^2 = 4R^2\sin^2 A$. Then …

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