Q.Find an angle $\theta$, $0 < \theta < \dfrac{\pi}{2}$, which increases twice as fast as its sine.

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The area of a triangle is obtained with lengths of two sides and included angle between them. If the angle is measured as $60^\circ 20'$ instead of $60^\circ$, then the percentage error in its area is (A) $\dfrac{5\pi}{54}$ (B) $\dfrac{5\pi}{27\sqrt{3}}$ (C) $\dfrac{5\sqrt{3}\pi}{27}$ (D) $\dfrac{5\pi}{27}$

   ›Reveal solutionSolution

   The percentage error in area due to a small angular error is found by differentiating the area formula $A = \frac12 ab\sin C$; the relative error is $\cot C \cdot dC$ (in radians). With $C=60^\circ$, $dC=20'=\frac{\pi}{540}$ rad, the percentage error becomes $\frac{5\pi}{27\sqrt{3}}$, matching option (B).

   Concept & Intuition

   When a quantity is computed from measured values, a small error in one measurement propagates into the result. Here, area $A = \frac12 ab \sin C$ depends on the included angle $C$. If the sides $a$ and $b$ are exact, the only source of error is the angle. For small errors, we use differentials: the change in area $dA \approx \frac{dA}{dC} \cdot dC$, and the relative error is $\frac{dA}{A} = \cot C \cdot dC$ (with $dC$ in radians). This turns a messy trigonometric problem into a simple calculus step.

   Step-by-step solution

   1. Write the exact area formula The area of a triangle with two sides $a$, $b$ and included angle $C$ is

   $A = \frac12 ab \sin C.$

   Here $a$ and $b$ are assumed error‑free; only $C$ is measured incorrectly.

   2. Find the differential of $A$ with respect to $C$ Differentiate:

   $\frac{dA}{dC} = \frac12 ab \cos C.$

   Hence a small error $dC$ in the angle causes an error in area:

   $dA \approx \frac12 ab \cos C \cdot dC.$

   3. Compute the relative error The relative error is

   $\frac{dA}{A} = \frac{\frac12 ab \cos C \cdot dC}{\frac12 ab \sin C} = \cot C \cdot dC.$

   This is the key formula: the relative error in area equals $\cot C$ times the angular error (in radians).

   4. Convert the angular error to radians The measured angle is $60^\circ 20'$ instead of $60^\circ$, so the error is

   $dC = 20' = \frac{20}{60}^\circ = \frac13^\circ.$

   Convert degrees to radians:

   $1^\circ = \frac{\pi}{180} \text{ rad} \quad\Rightarrow\quad dC = \frac13 \cdot \frac{\pi}{180} = \frac{\pi}{540} \text{ rad}.$

   5. Evaluate $\cot C$ at $C = 60^\circ$ …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the rate of change of volume of a cube and that of its surface area are numerically equal, then the length of its diagonal is (A) $2\sqrt{3}$ (B) $\sqrt{3}$ (C) $4\sqrt{3}$ (D) $6\sqrt{3}$

   ›Reveal solutionSolution

   Equating $\dfrac{dV}{dt}$ and $\dfrac{dS}{dt}$ numerically gives edge $x=4$, so the diagonal is $4\sqrt{3}$.

   Let the edge length be $x$. Then

   $V = x^3 \implies \frac{dV}{dt} = 3x^2\,\frac{dx}{dt},$

   $S = 6x^2 \implies \frac{dS}{dt} = 12x\,\frac{dx}{dt}.$

   Numerically equal:

   $3x^2 = 12x \implies x = 4.$ …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the parabola $y^2 = 9x$ cuts the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{b} = 1$ orthogonally, then the length of the latus rectum of the given parabola is (A) $2b$ (B) $\dfrac{b}{2}$ (C) $b$ (D) $\dfrac{b}{3}$

   ›Reveal solutionSolution

   Orthogonality forces $b=18$, so the parabola's latus rectum $= 9 = \dfrac{b}{2}$ — option (B).

   Setup. Parabola $y^2 = 9x$ (so $4a = 9$, latus rectum $= 4a = 9$) and ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{b} = 1$.

   Slopes at a common point.

   • Parabola: $2y\,y' = 9 \Rightarrow y'_p = \dfrac{9}{2y}$.
   • Ellipse: $\dfrac{2x}{9} + \dfrac{2y}{b}y' = 0 \Rightarrow y'_e = -\dfrac{bx}{9y}$.

   Orthogonality ($y'_p\, y'_e = -1$):

   $\frac{9}{2y}\cdot\left(-\frac{bx}{9y}\right) = -1 \;\Rightarrow\; \frac{bx}{2y^2} = 1 \;\Rightarrow\; bx = 2y^2 .$ …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the base of an isosceles triangle is $3\sqrt{2}$ feet and the two equal sides of it are increasing at the rate of $1$ ft/s, then the rate of increase of its area (in sq.ft/sec) when the angle between the equal sides is a right angle is (A) $3\sqrt{3}$ (B) $\sqrt{3}$ (C) $9$ (D) $3$

   ›Reveal solutionSolution

   The area of an isosceles triangle is expressed in terms of the equal side length and the included angle. Using the given rate of change of the side and the fact that the angle is fixed at the instant of interest, the rate of increase of area is found to be $3$ sq.ft/sec.

   The problem gives an isosceles triangle with base $3\sqrt{2}$ feet and equal sides that are increasing at $1$ ft/s. We need the rate of increase of its area at the moment when the angle between the equal sides is a right angle.

   The key is to choose a formula for area that directly involves the changing quantity (the equal side length) and the angle. For any triangle, area is $\frac12 ab \sin C$. Here, the two equal sides are the ones forming the included angle, so that formula is perfect.

   1. Set up the variables. Let the equal sides each have length $s$ feet, and let $\theta$ be the angle between them. The area $A$ of the triangle is

   $A = \frac12 \cdot s \cdot s \cdot \sin\theta = \frac12 s^2 \sin\theta.$

   2. What is given and what is wanted? We know $\frac{ds}{dt} = 1$ ft/s. We want $\frac{dA}{dt}$ at the instant when $\theta = 90^\circ = \frac{\pi}{2}$ radians. But note: the base is fixed at $3\sqrt{2}$ feet. Does that give a relation between $s$ and $\theta$? Yes — by the law of cosines, the base $b$ satisfies

   $b^2 = s^2 + s^2 - 2s^2 \cos\theta = 2s^2(1 - \cos\theta).$

   So $b = 3\sqrt{2}$ is constant, meaning $s$ and $\theta$ are not independent — as $s$ increases, $\theta$ must change to keep the base fixed. However, we only need the rate at a specific instant, not a full functional relation.

   3. Differentiate the area with respect to time. Since both $s$ and $\theta$ can change with time,

   $\frac{dA}{dt} = \frac12 \left( 2s \frac{ds}{dt} \sin\theta + s^2 \cos\theta \cdot \frac{d\theta}{dt} \right) = s \frac{ds}{dt} \sin\theta + \frac12 s^2 \cos\theta \frac{d\theta}{dt}.$

   4. Find $s$ and $\frac{d\theta}{dt}$ at the required instant. At $\theta = \frac{\pi}{2}$, $\sin\theta = 1$, $\cos\theta = 0$. The law of cosines gives

   $(3\sqrt{2})^2 = 2s^2 (1 - \cos\frac{\pi}{2}) = 2s^2 (1 - 0) = 2s^2.$

   So $18 = 2s^2$, hence $s^2 = 9$ and $s = 3$ feet (positive length).

   Now we need $\frac{d\theta}{dt}$ at that instant. Differentiate the law of cosines relation with respect to time. From $b^2 = 2s^2(1 - \cos\theta)$, since $b$ is constant,

   $0 = \frac{d}{dt}\left[2s^2(1 - \cos\theta)\right] = 4s \frac{ds}{dt}(1 - \cos\theta) + 2s^2 \sin\theta \frac{d\theta}{dt}.$

   At $\theta = \frac{\pi}{2}$, $\cos\theta = 0$, $\sin\theta = 1$, $s = 3$, $\frac{ds}{dt} = 1$: …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If P is any point on the curve $y^2 = 4ax$, other than the origin, then the length of the subtangent at P, y-coordinate of P and the length of the subnormal at P are in (A) arithmetic progression (B) arithmetic-Geometric progression (C) harmonic progression (D) geometric progression

   ›Reveal solutionSolution

   For a point $P$ on the parabola $y^2 = 4ax$, the subtangent, the y-coordinate, and the subnormal are in geometric progression. The correct option is (D).

   ---

   The key idea is to recall the geometric meanings of subtangent and subnormal for a curve. For a point $P$ on a curve, the subtangent is the projection of the tangent segment onto the x-axis, and the subnormal is the projection of the normal segment onto the x-axis. Their lengths have simple formulas in terms of the derivative.

   For any curve $y = f(x)$, at a point $(x, y)$:

   • Length of subtangent = $\left| \frac{y}{dy/dx} \right|$
   • Length of subnormal = $\left| y \cdot \frac{dy}{dx} \right|$

   Here the curve is given implicitly as $y^2 = 4ax$. We’ll use these formulas and then check the progression among the three quantities.

   ---

   1. Find the derivative. Differentiate $y^2 = 4ax$ with respect to $x$:

   $2y \frac{dy}{dx} = 4a \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2a}{y}.$

   2. Compute the subtangent.

      Subtangent length = $\left| \frac{y}{dy/dx} \right| = \left| \frac{y}{2a/y} \right| = \left| \frac{y^2}{2a} \right|$.

      Since $y^2 = 4ax$, this becomes $\left| \frac{4ax}{2a} \right| = 2|x|$.

      For a point on the parabola (other than the origin), $x > 0$ (right-opening parabola), so subtangent = $2x$.

   3. The y-coordinate of P is simply $y$.

   4. Compute the subnormal.

      Subnormal length = $\left| y \cdot \frac{dy}{dx} \right| = \left| y \cdot \frac{2a}{y} \right| = 2a$.

      Notice this is constant — independent of the point $P$! That’s a neat property of the parabola.

   5. Now we have three numbers:

      • Subtangent: $2x$
      • y-coordinate: $y$
      • Subnormal: $2a$

      We need to check if they are in arithmetic, geometric, or harmonic progression. …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The slope of a common tangent to the circles $x^2 + y^2 = 16$ and $(x-9)^2 + y^2 = 16$ is (A) $\dfrac{8}{\sqrt{13}}$ (B) $\dfrac{4}{\sqrt{13}}$ (C) $\dfrac{\sqrt{17}}{8}$ (D) $\dfrac{8}{\sqrt{17}}$

   ›Reveal solutionSolution

   The common tangent to two equal circles is found by considering the line’s distance from each centre equals the radius; solving gives slope $\pm \frac{8}{\sqrt{17}}$, so the correct option is (D).

   We have two circles of equal radius $4$:

   • Circle 1: centre $O_1 = (0,0)$, radius $r = 4$
   • Circle 2: centre $O_2 = (9,0)$, radius $r = 4$

   A common tangent touches both circles. Because the circles are the same size, the common tangents are either direct (parallel to the line joining centres) or transverse (crossing between them). Here we want a common tangent — the slope will be the same for both points of tangency.

   ---

   Why this approach works

   For any line to be tangent to a circle, the perpendicular distance from the circle’s centre to the line must equal the radius. If the same line is tangent to both circles, then the distances from $O_1$ and $O_2$ to the line are both $4$. This gives two equations in the line’s parameters, which we can solve for the slope.

   ---

   Step-by-step solution

   1. Write the general line equation Let the common tangent have slope $m$ and intercept $c$:

   $y = mx + c \quad \Rightarrow \quad mx - y + c = 0$

   2. Distance from centre $(0,0)$ to the line equals radius $4$

   $\frac{|m\cdot 0 - 0 + c|}{\sqrt{m^2 + 1}} = 4 \quad \Rightarrow \quad \frac{|c|}{\sqrt{m^2+1}} = 4 \qquad(1)$

   3. Distance from centre $(9,0)$ to the same line also equals $4$

   $\frac{|m\cdot 9 - 0 + c|}{\sqrt{m^2+1}} = 4 \quad \Rightarrow \quad \frac{|9m + c|}{\sqrt{m^2+1}} = 4 \qquad(2)$

   4. Equate the two distances From (1) and (2):

   $|c| = |9m + c|$

   This gives two cases:

   • Case 1: $c = 9m + c \quad \Rightarrow \quad 9m = 0 \quad \Rightarrow \quad m = 0$

     Then from (1): $\frac{|c|}{1} = 4 \Rightarrow c = \pm 4$.

     This gives horizontal tangents $y = \pm 4$ — these are indeed common tangents (top and bottom), but slope $0$ is not among the options.

   • Case 2: $c = -(9m + c) \quad \Rightarrow \quad 2c = -9m \quad \Rightarrow \quad c = -\frac{9m}{2}$

   5. Substitute $c$ into the distance condition (1) …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The locus of a point which divides the line segment joining the focus and any point on the parabola $y^2 = 12x$ in the ratio $m:n$ $(m+n \neq 0)$ is a parabola. Then the length of the latus rectum of that parabola is (A) $\dfrac{m}{m+n}$ (B) $\dfrac{12m}{m+n}$ (C) $\dfrac{m}{12(m+n)}$ (D) $\dfrac{n}{12(m+n)}$

   ›Reveal solutionSolution

   The locus of the point dividing the segment from the focus to a point on the parabola in a fixed ratio is itself a parabola; its latus rectum is $\frac{12m}{m+n}$, so the correct option is (B).

   We start with the parabola $y^2 = 12x$. Its standard form is $y^2 = 4ax$, so $4a = 12$ gives $a = 3$. The focus is at $(a,0) = (3,0)$.

   Let $P(t)$ be any point on the parabola. Using the parametric form $x = at^2$, $y = 2at$, with $a=3$, we have

   $P = (3t^2,\ 6t)$.

   Let $Q$ be the point that divides the segment joining the focus $F(3,0)$ and $P$ in the ratio $m:n$, with $m$ corresponding to the segment from $F$ to $Q$ and $n$ from $Q$ to $P$ (the order matters for the section formula). Then by the section formula:

   $Q = \left( \frac{n \cdot 3 + m \cdot 3t^2}{m+n},\ \frac{n \cdot 0 + m \cdot 6t}{m+n} \right) = \left( \frac{3n + 3m t^2}{m+n},\ \frac{6m t}{m+n} \right).$

   We want the locus of $Q$ as $t$ varies. Let the coordinates of $Q$ be $(X,Y)$. Then:

   $X = \frac{3n + 3m t^2}{m+n}, \qquad Y = \frac{6m t}{m+n}.$

   From the expression for $Y$, solve for $t$:

   $t = \frac{(m+n)Y}{6m}.$

   Substitute into $X$:

   $X = \frac{3n + 3m \left( \frac{(m+n)Y}{6m} \right)^2}{m+n} = \frac{3n + 3m \cdot \frac{(m+n)^2 Y^2}{36 m^2}}{m+n} = \frac{3n + \frac{(m+n)^2 Y^2}{12 m}}{m+n}.$

   Multiply numerator and denominator:

   $X = \frac{3n}{m+n} + \frac{(m+n) Y^2}{12 m}.$

   Rearrange to isolate $Y^2$:

   $$X - \frac{3n}{m+n} = \frac{(m+n) Y^2}{12 m} \quad\Rightarrow\quad …$$

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Two ships leave a port at the same time. One of them moves in the direction of E$50^\circ$N with a speed of 8 kmph and the other moves in the direction of S$20^\circ$E with a speed of 12 kmph. Then the distance between the ships at the end of 2 hours is (in km) (A) $8\sqrt{7}$ (B) $34$ (C) $8\sqrt{19}$ (D) $32$

   ›Reveal solutionSolution

   The distance between the ships is $8\sqrt{19}$ km, so the answer is (C).

   In $2$ hours the ships travel $OA=8\times2=16$ km and $OB=12\times2=24$ km from the port $O$.

   Angle between the two paths (as bearings from north).

   • E$50^\circ$N $=$ N$40^\circ$E $\Rightarrow$ bearing $40^\circ$.
   • S$20^\circ$E $\Rightarrow$ bearing $180^\circ-20^\circ=160^\circ$.

   So $\angle AOB = 160^\circ-40^\circ = 120^\circ$.

   Law of cosines. …

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9. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of K miles/hour. If the rate of increase of his shadow is $\dfrac{11}{5}$ feet/sec, then K = (Take 1 mile = 5280 feet) (A) 2 (B) 3 (C) 4 (D) 5

   ›Reveal solutionSolution

   Using similar triangles the shadow length is $s=\tfrac{x}{2}$, so $\tfrac{ds}{dt}=\tfrac12\tfrac{dx}{dt}$. From $\tfrac{ds}{dt}=\tfrac{11}{5}$ ft/s the man's speed is $\tfrac{22}{5}$ ft/s $=3$ mph. Answer: (B) 3.

   Setup (similar triangles). Let the lamp be at height $15$ ft, the man ($5$ ft tall) at distance $x$ ft from the lamp post, and $s$ the length of his shadow. The lamp-ground-shadow-tip triangle and the man-feet-shadow-tip triangle are similar:

   $\frac{15}{x+s}=\frac{5}{s}\;\Rightarrow\;15s=5x+5s\;\Rightarrow\;10s=5x\;\Rightarrow\;s=\frac{x}{2}.$

   Differentiate.

   $\frac{ds}{dt}=\frac12\,\frac{dx}{dt}.$

   Solve for the man's speed. Given $\dfrac{ds}{dt}=\dfrac{11}{5}$ ft/s,

   $\frac{dx}{dt}=2\cdot\frac{11}{5}=\frac{22}{5}\ \text{ft/s}.$ …

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10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.There is a possible error of 0.03 cm in a scale of length 1 foot with which the height of a closed right circular cylinder and the diameter of a sphere are measured as 3.5 feet each. If the radii of both cylinder and sphere are same, then the approximate error in the sum of the surface areas of both cylinder and sphere is (in square feet) (A) 0.385 (B) 0.0962 (C) 0.77 (D) 0.1925

    ›Reveal solutionSolution

    With equal radii, the total surface area is $S=\pi d h+\tfrac32\pi d^2$; propagating the scale error by differentials gives $\Delta S=17.5\pi\,\Delta x\approx0.1925$ ft$^2$. Answer: (D) 0.1925.

    Surface areas (radii equal, cylinder closed). Let the common radius be $r=\tfrac{d}{2}$ with measured diameter $d=3.5$ ft and height $h=3.5$ ft.

    $S=\underbrace{(2\pi r h+2\pi r^2)}_{\text{closed cylinder}}+\underbrace{4\pi r^2}_{\text{sphere}}=2\pi r h+6\pi r^2.$

    In terms of the measured quantities $d$ and $h$ (with $r=d/2$):

    $S=\pi d h+\frac{3}{2}\pi d^2.$

    Differential error.

    $dS=\frac{\partial S}{\partial d}\,\Delta d+\frac{\partial S}{\partial h}\,\Delta h,\qquad \frac{\partial S}{\partial d}=\pi h+3\pi d,\quad \frac{\partial S}{\partial h}=\pi d.$

    At $d=h=3.5$:

    $\frac{\partial S}{\partial d}=\pi(3.5)+3\pi(3.5)=14\pi,\qquad \frac{\partial S}{\partial h}=3.5\pi.$ …

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11. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of K miles/hour. If the rate of increase of his shadow is $\dfrac{11}{5}$ feet/sec, then K = (Take 1 mile = 5280 feet) (A) 2 (B) 3 (C) 5 (D) 4

    ›Reveal solutionSolution

    $K = 3$ miles/hour — option (B).

    By similar triangles, if $x$ is the man's distance from the pole and $s$ his shadow's length, the tip of the shadow, the top of the lamp and the man's head are collinear:

    $\frac{x+s}{15} = \frac{s}{5} \;\Rightarrow\; 5(x+s) = 15s \;\Rightarrow\; 5x = 10s \;\Rightarrow\; s = \frac{x}{2}.$

    Differentiating with respect to time:

    $\frac{ds}{dt} = \frac{1}{2}\,\frac{dx}{dt}.$

    Given $\dfrac{ds}{dt} = \dfrac{11}{5}$ ft/sec, …

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.Two ships leave a port at the same time. One of them moves in the direction of E$50^\circ$N with a speed of 8 kmph and the other moves in the direction of S$20^\circ$E with a speed of 12 kmph. Then the distance between the ships at the end of 2 hours is (in km) (A) 34 (B) 32 (C) $8\sqrt{7}$ (D) $8\sqrt{19}$

    ›Reveal solutionSolution

    The problem is a classic relative-motion / vector-addition problem: find the distance between two ships after 2 hours by computing the vector difference of their displacements. The correct answer is $8\sqrt{19}$ km, which corresponds to option (D).

    We have two ships leaving the same port at the same time, each moving along a different bearing. To find the distance between them after a given time, we can treat their paths as vectors from the origin (the port). The distance between them is the magnitude of the difference of these two vectors.

    Why this works:

    Instead of drawing a triangle and using the law of cosines directly (which is also valid), vector subtraction neatly handles the directions. Each ship’s displacement is simply (speed × time) in its given direction. The difference vector’s length is the straight-line separation.

    ---

    1. Convert bearings to standard angles (measured from the positive x‑axis, i.e., East).

       • Ship A: direction E $50^\circ$ N means $50^\circ$ north of east. So its angle from East is $+50^\circ$ (counterclockwise). Standard angle: $\theta_A = 50^\circ$.
       • Ship B: direction S $20^\circ$ E means $20^\circ$ east of south. South is $270^\circ$ (or $-90^\circ$), so adding $20^\circ$ eastward gives $270^\circ + 20^\circ = 290^\circ$ (or equivalently $-70^\circ$). Standard angle: $\theta_B = 290^\circ$.

    2. Compute displacement vectors after 2 hours.

       Speed of A = 8 km/h → distance = $8 \times 2 = 16$ km.

       Speed of B = 12 km/h → distance = $12 \times 2 = 24$ km.

       Vector for A:

    $\vec{A} = 16\,(\cos 50^\circ,\ \sin 50^\circ)$

    Vector for B:

    $\vec{B} = 24\,(\cos 290^\circ,\ \sin 290^\circ)$

    Since $\cos 290^\circ = \cos(360^\circ-70^\circ) = \cos 70^\circ$ and $\sin 290^\circ = -\sin 70^\circ$, we have:

    $\vec{B} = 24\,(\cos 70^\circ,\ -\sin 70^\circ)$

    3. Find the vector from ship A to ship B (or vice versa).

    $\vec{D} = \vec{B} - \vec{A} = (24\cos 70^\circ - 16\cos 50^\circ,\ -24\sin 70^\circ - 16\sin 50^\circ)$

    4. Compute the squared distance $|\vec{D}|^2$.

    $|\vec{D}|^2 = (24\cos 70^\circ - 16\cos 50^\circ)^2 + (-24\sin 70^\circ - 16\sin 50^\circ)^2$

    Expand:

    $= 24^2\cos^2 70^\circ + 16^2\cos^2 50^\circ - 2\cdot24\cdot16\cos70^\circ\cos50^\circ$

    $+ 24^2\sin^2 70^\circ + 16^2\sin^2 50^\circ + 2\cdot24\cdot16\sin70^\circ\sin50^\circ$

    Notice $\cos^2\theta + \sin^2\theta = 1$ for each angle, so:

    $= 24^2 + 16^2 + 2\cdot24\cdot16\,(\sin70^\circ\sin50^\circ - \cos70^\circ\cos50^\circ)$ …

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