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Properties-and-solution-of-triangle Formula Sheet

This page will help you to revise formulas and concepts of Properties-and-solution-of-triangle instantly for various exams.
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Inverse trigonometric functions reverse the usual ratios, allowing you to find angles when given a ratio. They include functions like arcsin, arccos, and arctan, which map values from the ratio back to an angle in specific ranges. Think of them as the GPS that gets you back to the angle when you’ve wandered too far into sine, cosine, or tangent territory!

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Neetesh Kumar

Neetesh Kumar | September 06, 2024                                       \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space Share this Page on: Reddit icon Discord icon Email icon WhatsApp icon Telegram icon



1. Sine Formula:

    In any triangle, ABC

      asin A=bsin B=csin C=λ=abc2Δ=2R\space \space \space \space \space \space \frac{a}{sin \space A} = \frac{b}{sin \space B} = \frac{c}{sin \space C} = \lambda = \frac{abc}{2\Delta} = 2R

     \space \space \space \space \space where R is circumradius and Δ\Delta is area of triangle.

2. Cosine Formula:

      \space \space \space \space \space \space (a) cos A=b2+c2a22bc or a2=b2+c22bc cosAcos \space A = \frac{b^2+c^2-a^2}{2bc} \space or \space a^2 = b^2 + c^2 - 2bc \space cos A

      \space \space \space \space \space \space (b) cos B=c2+a2b22cacos \space B = \frac{c^2+a^2-b^2}{2ca}

      \space \space \space \space \space \space (c) cos C=a2+b2c22abcos \space C = \frac{a^2+b^2-c^2}{2ab}

3. Projection Formula:

      \space \space \space \space \space \space (a) b cos C+c cos B=ab \space cos \space C + c \space cos \space B = a

      \space \space \space \space \space \space (b) c cos A+a cos C=bc \space cos \space A + a \space cos \space C = b

      \space \space \space \space \space \space (c) a cos B+b cos A=ca \space cos \space B + b \space cos \space A = c

4. Napier's Analogy (Tangent Rule):

      \space \space \space \space \space \space (a) tanBC2=bcb+ccotA2tan \big\lgroup \frac{B-C}{2} \big\rgroup = \frac{b-c}{b+c} cot \frac{A}{2}

      \space \space \space \space \space \space (b) tanCA2=cac+acotB2tan \big\lgroup \frac{C-A}{2} \big\rgroup = \frac{c-a}{c+a} cot \frac{B}{2}

      \space \space \space \space \space \space (c) tanAB2=aba+bcotC2tan \big\lgroup \frac{A-B}{2} \big\rgroup = \frac{a-b}{a+b} cot \frac{C}{2}

5. Half Angle Formula:

            s=a+b+c2 \space \space \space \space \space \space \space \space \space \space \space \space s = \frac{a+b+c}{2} = semi-perimeter of triangle.

      \space \space \space \space \space \space (a) (i) sinA2=(sb)(sc)bc     sin \frac{A}{2} = \sqrt \frac{(s-b)(s-c)}{bc} \space \space \space \space \space (ii) sinB2=(sc)(sa)casin \frac{B}{2} = \sqrt \frac{(s-c)(s-a)}{ca}

           \space \space \space \space \space \space \space \space \space \space \space (iii) sinC2=(sa)(sb)absin \frac{C}{2} = \sqrt \frac{(s-a)(s-b)}{ab}

       \space \space \space \space \space \space \space (b) (i) cosA2=s(sa)bc          cos \frac{A}{2} = \sqrt \frac{s(s-a)}{bc} \space \space \space \space \space \space \space \space \space \space (ii) cosB2=s(sb)cacos \frac{B}{2} = \sqrt \frac{s(s-b)}{ca}

            \space \space \space \space \space \space \space \space \space \space \space \space (iii) cosC2=s(sc)abcos \frac{C}{2} = \sqrt \frac{s(s-c)}{ab}

       \space \space \space \space \space \space \space (c) (i) tanA2=(sb)(sc)s(sa)=Δs(sa)tan \frac{A}{2} = \sqrt \frac{(s-b)(s-c)}{s(s-a)} = \frac{\Delta}{s(s-a)}

            \space \space \space \space \space \space \space \space \space \space \space \space (ii) tanB2=(sc)(sa)s(sb)=Δs(sb)tan \frac{B}{2} = \sqrt \frac{(s-c)(s-a)}{s(s-b)} = \frac{\Delta}{s(s-b)}

            \space \space \space \space \space \space \space \space \space \space \space \space (iii) tanc2=(sa)(sb)s(sc)=Δs(sc)tan \frac{c}{2} = \sqrt \frac{(s-a)(s-b)}{s(s-c)} = \frac{\Delta}{s(s-c)}

       \space \space \space \space \space \space \space(d) Area of Triangle

            Δ=s(sa)(sb)(sc)\space \space \space \space \space \space \space \space \space \space \space \space \Delta = \sqrt{s(s-a)(s-b)(s-c)}

        =12bc sin A=12ca sin B=12ab sin C\space \space \space \space \space \space \space \space =\frac{1}{2}bc \space sin \space A = \frac{1}{2}ca \space sin \space B = \frac{1}{2}ab \space sin \space C

                 =142(a2b2+b2c2+c2a2)a4b4c4\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space = \frac{1}{4}\sqrt{2({a^2}{b^2}+{b^2}{c^2}+{c^2}{a^2})-{a^4}-{b^4}-{c^4}}

6. Radius of the Circumcircle 'R':

The circumcentre is the point of intersection of perpendicular bisectors of the sides, and the distance between the circumcentre & vertex of the triangle is called circumradius 'R'.

      R=a2 sin A=b2 sin B=c2 sin C=abc4Δ.\space \space \space \space \space \space R = \frac{a}{2 \space sin \space A} = \frac{b}{2 \space sin \space B} = \frac{c}{2 \space sin \space C} = \frac{abc}{4\Delta}.

7. The radius of the Incircle 'r':

The point of intersection of internal angle bisectors is the incentre, and the perpendicular distance of the incentre from any side is called inradius 'r'.

        r=Δs=(sa)tanA2=(sb)tanB2\space \space \space \space \space \space \space \space r = \frac{\Delta}{s} = (s-a) tan \frac{A}{2} = (s-b) tan \frac{B}{2}

       =(sc)tanC2=4R sinA2sinB2sinC2.\space \space \space \space \space \space \space =(s-c) tan \frac{C}{2} = 4R \space sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}.

8. Radii of the Ex-Circles:

The point of intersection of two external angles and one internal angle bisector is the excentre, and the perpendicular distance of the excentre from any side is called an exradius. If r1{r_1} is the radius of the escribed circle opposite to angle A of ΔABC{\Delta}ABC and so on, then :

     \space \space \space \space \space (a) r1=Δsa=s tanA2=4R sinA2cosB2cosC2r_1 = \frac {\Delta}{s-a} = s \space tan \frac{A}{2} = 4R \ sin \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

     \space \space \space \space \space (b) r2=Δsb=s tanB2=4R cosA2sinB2cosC2r_2 = \frac {\Delta}{s-b} = s \space tan \frac{B}{2} = 4R \ cos \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2}

     \space \space \space \space \space (c) r3=Δsc=s tanC2=4R cosA2cosB2sinC2r_3 = \frac {\Delta}{s-c} = s \space tan \frac{C}{2} = 4R \space cos \frac{A}{2} cos \frac{B}{2} sin \frac{C}{2}

9. Length of Angle Bisector, Median & Altitude:

if ma,βa & ha{m_a},{\beta_a} \space \& \space {h_a} are the lengths of a median, an angle bisector & altitude from the angle A then,

                12b2+c2+2bc cosA=ma=122b2+2c22a2\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \frac{1}{2} \sqrt {b^2+c^2+2bc \ cos A} = {m_a} = \frac{1}{2} \sqrt {2b^2+2c^2-2a^2}

      \space \space \space \space \space \space and   \space \space βa=2bc cosA2b+c,          \beta{_a} = \frac{2bc \space cos \frac{A}{2}}{b+c}, \space \space \space \space \space \space \space \space \space \space ha=acotB+cotC{h_a} = \frac{a}{cot B + cot C}

      \space \space \space \space \space \space Note that ma2+mb2+mc2=34(a2+b2+c2)m_a^2 + m_b^2 + m_c^2 = \frac{3}{4} (a^2+b^2+c^2)

10. Orthocentre and Orthic Triangle:

(a) The point of intersection of altitudes is orthocentre & the triangle KLM, which is formed by joining the feet of the altitudes, is called the orthic triangle.

(b) The distances of the orthocentre from the angular points of the ΔABC{\Delta}{ABC} are 2R cosA,2R cosB2R \space cos A, 2R \space cos B & 2R cosC.2R \space cos C.

(c) The distance of orthocentre from sides are 2R cosB cosC, 2R cosC cosA2R \space cos B \space cos C, \space 2R \space cos C \space cos A and 2R cosA cosB2R \space cos A \space cos B

(d) The sides of the orthic triangle are a cosA(=R sin2A),b cosB(=R sin2B)a \space cos A ( = R \space sin 2A), b \space cos B (= R \space sin 2B) and c cosC(=R sin2C)c \space cos C (=R \space sin 2C) and its angles are π2A,π2B\pi - 2A, \pi - 2B and π2C\pi - 2C

(e) Circumradii of the triangles PBC,PCA,PABPBC, PCA, PAB and ABCABC are equal.

(f) Area of orthic triangle = 2Δ cosA cosB cosC=12R2 sin2A sin2B sin2C2\Delta \space cosA \space cosB \space cosC = \frac{1}{2}R^2 \space sin2A \space sin2B \space sin2C

(g) Circumradii of orthic triangle =R2= \frac{R}{2}

11. Ex-Central Triangle:

(a) The triangle formed by joining the three excentres I1,I2{\Iota_1}, {\Iota_2} and I3{\Iota_3} Of ΔABC\Delta ABC is called the excentral or excentric triangle.

(b) Incentre I{\Iota} of ΔABC\Delta ABC is the orthocentre of the excentral ΔI1I2I3.\Delta {\Iota_1} {\Iota_2}{\Iota_3}.

(C) ΔABC\Delta ABC is the orthic triangle of the ΔI1I2I3.\Delta {\Iota_1} {\Iota_2}{\Iota_3}.

(d) The sides of the excentral triangle are 4R cosA2,4R cosB24R \space cos \frac{A}{2}, 4R \space cos \frac{B}{2} and 4R cosC24R \space cos \frac{C}{2} and its angles are π2A2,π2B2\frac{\pi}{2} - \frac{A}{2}, \frac{\pi}{2} - \frac{B}{2} and π2C2.\frac{\pi}{2} - \frac{C}{2}.

(e) II1=4R sinA2;II2=4R sinB2;II3=4R sinC2.{\Iota \Iota_1} = 4R \space sin \frac{A}{2}; {\Iota \Iota_2} = 4R \space sin \frac{B}{2}; {\Iota \Iota_3} = 4R \space sin \frac{C}{2}.

12. The Distance between the Special Points:

(a) The distance between circumcentre and orthocentre   =R18cosAcosBcosC\space \space = R \sqrt{1 -8 cos A cos B cos C}

(b) The distance between circumcentre and incentre =R22Rr= \sqrt{R^2 - 2Rr}

(c) The distance between incentre and orthocentre   =2R24R2 cosAcosBcosC\space \space = \sqrt{2R^2 - 4R^2 \space cos A cos B cos C}

(d) The distance between circumcentre & excentre is
OI1=R1+8sinA2cosB2cosC2=R2+2R1\Omicron \Iota _1 = R \sqrt{1+8 sin \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}} = \sqrt {R^2 + 2R}{_1} & so on.

13. m-n Theorem:

        (m+n)cot θ=m cot an cot β\space \space \space \space \space \space \space \space (m + n) cot \space \theta = m \space cot \space a - n \space cot \space \beta
        (m+n)cot θ=n cot Bm cot C.\space \space \space \space \space \space \space \space(m + n) cot \space \theta = n \space cot \space B - m \space cot \space C.

14. Important Points:

       \space \space \space \space \space \space \space (a) (i) If a cos B=b cos A,a \space cos \space B = b \space cos \space A, then the triangle is isosceles.
            \space \space \space \space \space \space \space \space \space \space \space \space (ii) If a cos A=b cos B,a \space cos \space A = b \space cos \space B, then the triangle is isosceles or right angled.

       \space \space \space \space \space \space \space (b) In Right Angle Triangle :

             \space \space \space \space \space \space \space \space \space \space \space \space \space (i) a2+b2+c2=8R2a^2 + b^2 + c^2 = 8R^2

             \space \space \space \space \space \space \space \space \space \space \space \space \space (ii) cos2A+cos2B+cos2C=1cos^2 A + cos^2 B + cos^2 C = 1

       \space \space \space \space \space \space \space (c) In equilateral triangle :

            \space \space \space \space \space \space \space \space \space \space \space \space (i) R=2rR = 2r                             \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space (ii) r1=r2=r3=3R2r_1 = r_2 = r_3 = \frac{3R}{2}

            \space \space \space \space \space \space \space \space \space \space \space \space (iii) r:R:r1=1:2:3r : R : r_1 = 1 : 2 : 3      \space \space \space \space \space (iv) area=3a24area = \frac{\sqrt3{a^2}}{4}      \space \space \space \space \space (v) R=a3R = \frac{a}{\sqrt3}

(d) (i) The circumcentre lies (1) inside an acute angled triangle (2) outside an obtuse angled triangle & (3) at mid point of the hypotenuse of right angled triangle

(ii) The orthocentre of right angled triangle is the vertex at the right angle.

(iii) The orthocentre, centroid & circumcentre are collinear & centroid divides the line segment joining orthocentre & circumcentre internally in the ratio 2 : 1, except in case of equilateral triangle. In equilateral triangle all these centres coincide.

15. Regular Polygon:

Consider a 'n' sided regular polygon of side length 'a'

(a) Radius of incircle of this polygon r=a2cotπnr = \frac{a}{2} cot \frac{\pi}{n}

(b) Radius of circumcircle of this polygon R=a2cosecπnR = \frac{a}{2} cosec \frac{\pi}{n}

(c) Perimeter & area of regular polygon Perimeter =na=2nr tanπn=2nR sinπn= na = 2nr \space tan \frac{\pi}{n} = 2nR \space sin \frac{\pi}{n} Area =12nR2sin2πn=nr2tanπn=14na2cotπn= \frac{1}{2} n R^2 sin \frac{2\pi}{n} = nr^2 tan \frac{\pi}{n} = \frac{1}{4} na^2 cot \frac{\pi}{n}

16. Cyclic Quadrilateral:

(a) Quadrilateral ABCDABCD is cyclic if A+C=π=B+C\angle A + \angle C = \pi = \angle B + \angle C (opposite angle are supplementary angles)

(b) Area =(sa)(sb)(sc)(sd),= \sqrt{(s-a)(s-b)(s-c)(s-d)}, where 2s=a+b+c+d2s = a+b+c+d

(c) cosB=a2+b2c2d22(ab+cd)cos B = \frac{a^2+b^2-c^2-d^2}{2(ab+cd)} & similarly other angles

(d) Ptolemy's theorem : If ABCDABCD is cyclic quadrilateral, then AC.BD=AB.CD+BC.ADAC . BD = AB . CD + BC . AD

17. Solution of Triangle:

Case-I: Three sides are given then to find out three angles use cosA=b2+c2a22bc, cosB=c2+a2b22ac, & cosC=a2+b2c22ab,cos A = \frac{b^2+c^2-a^2}{2bc}, \space cos B = \frac{c^2+a^2-b^2}{2ac}, \space \& \space cos C = \frac{a^2+b^2-c^2}{2ab},

Case-II : Two sides & included angle are given : Let sides a, b & angle C are given then use tanAB2=aba+bcotC2tan \frac{A-B}{2} = \frac{a-b}{a+b} cot \frac{C}{2} and find value of ABA-B ......(i)
& A+B2=90°C2\& \space \frac{A+B}{2} = 90 \degree - \frac{C}{2} ......(ii) c=a sin Csin Ac= \frac{a \space sin \space C}{sin \space A} ......(iii)

Case-III : Two sides a, b & angle A opposite to one of them is given

(a) If a<b sinAa < b \space sin A No triangle exist

(b) if a=b sinA & Aa = b \space sin A \space \& \space A is acute, then one triangle exist which is right angled.

(c) a>b sinA,a<b &Aa > b \space sin A, a < b \space \& A is acute, then two triangles exist

(d) a>b sinA,a>b & Aa > b \space sin A, a > b \space \& \space A is acute, then one triangle exist

(e) a>b sinA, & Aa > b \space sin A, \space \& \space A is obtuse, then there is one triangle if a>ba > b & no triangle if a<b. a < b.

Note: Case-III can be analyzed algebraically using the Cosine rule as cosA=b2+c2a22bc,cos A = \frac{b^2+c^2-a^2}{2bc}, which is quadratic in c.

18. Angles of Elevation and Depression:

Let OP be a horizontal line in the vertical plane in which an object R is given, and let OR be joined.

In Fig. (a), where the object R is above the horizontal line OP, the angle POR is called the angle of elevation of the object R as seen from the point O. In Fig. (b), where the object R is below the horizontal line OP, the angle POR is called the angle of depression of the object R, as seen from the point O.

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