Solutions of Triangles | Trigonometry
- Sine Formula:
In any triangle ABC, \(\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}\) - Cosine Formula:
(i) \(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\) or a² = b² + c² − 2bc. cos A
(ii) \(\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}\)
(iii) \(\cos \mathrm{C}=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2 \mathrm{ab}}\) - Projection Formula:
(i) a = b cos C + c cos B
(ii) b = c cos A + a cos C
(iii) c = a cos B + b cos A - Napier’S Analogy − Tangent Rule:
(i) \(\tan \frac{\mathrm{B}-\mathrm{C}}{2}=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \cot \frac{\mathrm{A}}{2}\)
(ii) \(\tan \frac{\mathrm{C}-\mathrm{A}}{2}=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{c}+\mathrm{a}} \cot \frac{\mathrm{B}}{2}\)
(iii) \(\tan \frac{\mathrm{A}-\mathrm{B}}{2}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}} \cot \frac{\mathrm{C}}{2}\) - Trigonometric Functions Of Half Angles:
(i) \(\sin \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{bc}}} ; \sin \frac{\mathrm{B}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{c})(\mathrm{s}-\mathrm{a})}{\mathrm{ca}}} ; \sin \frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}{\mathrm{ab}}}\)
(ii) \(\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} ; \cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{c a}} ; \cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{a b}}\)
(iii) \(\tan \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{s}(\mathrm{s}-\mathrm{a})}}=\frac{\Delta}{\mathrm{s}(\mathrm{s}-\mathrm{a})} \quad \text { where } \mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\) & ∆ = area of triangle.
(iv) Area of triangle = \(=\sqrt{s(s-a)(s-b)(s-c)}\) - M − N Rule:
In any triangle,
(m + n) cot θ = m cot α – n cot β
= n cot B m cot C - \(\frac{1}{2} ab \sin C=\frac{1}{2} bc \sin A=\frac{1}{2}ca\sin B =\) area of triangle ABC.
\(\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=2 \mathrm{R}\)
Note that \(\mathrm{R}=\frac{\mathrm{a} \mathrm{b} \mathrm{c}}{4 \Delta}\) ; Where R is the radius of circumcircle & ∆ is area of triangle - Radius of the incircle ‘r’ is given by:
(a) \(r=\frac{\Delta}{s} \text { where } s=\frac{a+b+c}{2}\)
(b) \(r=(s-a) \tan \frac{A}{2}=(s-b) \tan \frac{B}{2}=(s-c)\tan \frac{\mathrm{C}}{2}\)
(c) \(r=\frac{a \sin \frac{B}{2} \sin \frac{c}{2}}{\cos \frac{A}{2}} \& \text {so on}\)
(d) \(r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\) - Radius of the Ex− circles r1 , r2 & r3 are given by:
(a) \(r_{1}=\frac{\Delta}{s-a} ; r_{2}=\frac{\Delta}{s-b} ; r_{3}=\frac{\Delta}{s-c}\)
(b) \(r_{1}=s \tan \frac{A}{2} ; \quad r_{2}=s \tan \frac{B}{2} ; r_{3}=s \tan \frac{C}{2}\)
(c) \(r_{1}=\frac{a \cos \frac{B}{2} \cos \frac{c}{2}}{\cos \frac{A}{2}}\quad \& \text {so on}\)
(d) \(r_{1}=4 R \sin \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\)
\(\mathrm{r}_{2}=4 \mathrm{R} \sin \frac{\mathrm{B}}{2} \cdot \cos \frac{\mathrm{A}}{2} \quad . \cos \frac{\mathrm{C}}{2} \quad ; \quad \mathrm{r}_{3}=4 \mathrm{R} \sin \frac{\mathrm{C}}{2} \cdot \cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2}\) - Length of Angle Bisector & Medians:
If ma and βa are the lengths of a median and an angle bisector from the angle A then,
\(\mathrm{m}_{\mathrm{a}}=\frac{1}{2} \sqrt{2 \mathrm{b}^{2}+2 \mathrm{c}^{2}-\mathrm{a}^{2}} \text {and}\beta_{\mathrm{a}}=\frac{2 \mathrm{bc} \cos \frac{\mathrm{A}}{2}}{\mathrm{b}+\mathrm{c}}\)
Note that
\(m_{2}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)\) - Orthocentre And Pedal Triangle:
The triangle KLM which is formed by joining the feet of the altitudes is called the pedal triangle.−the distances ofthe orthocentre from the angular points ofthe∆ ABC are 2 R cosA , 2 R cosB and 2 R cos− the distances of P from sides are 2 R cosB cosC, 2 R cosC cosA and 2 R cosA cosB− the sides of the pedal triangle are a cosA (= R sin 2A), b cosB (= R sin 2B) and c cosC (= R sin 2C) and its angles areπ − 2A, π − 2B and π − 2C.
− circumradii of the triangles PBC, PCA, PAB and ABC are equal. - Excentral Triangle: The triangle formed by joining the three excentres I1, I2 and I3 of ∆ ABC is called the excentral or excentric triangle.
Note that: Incentre I of ∆ ABC is the orthocentre of the excentral ∆ I1I2I3.
− ∆ ABC is the pedal triangle of the ∆ I1I2I3.− the sides of the excentral triangle are
\(4 \mathrm{R} \cos \frac{\mathrm{A}}{2}, 4 \mathrm{R} \cos \frac{\mathrm{B}}{2} \text { and } 4 \mathrm{R} \cos \frac{\mathrm{C}}{2} \quad \text { and its angles are } \frac{\pi}{2}-\frac{\mathrm{A}}{2}, \frac{\pi}{2}-\frac{\mathrm{B}}{2} \text { and } \frac{\pi}{2}-\frac{\mathrm{C}}{2}\)
– \(\mathrm{II}_{1}=4 \mathrm{R} \sin \frac{\mathrm{A}}{2} ; \mathrm{II}_{2}=4 \mathrm{R} \sin \frac{\mathrm{B}}{2} ; \mathrm{II}_{3}=4 \mathrm{R} \sin \frac{\mathrm{C}}{2}\) - The Distances Between The Special Points:
(a) The distance between circumcentre and orthocentre is
\(=R \cdot \sqrt{1-8 \cos A \cos B \cos C}\)
(b) The distance between circumcentre and incentre is
\(=\sqrt{\mathrm{R}^{2}-2 \mathrm{R} \mathrm{r}}\)
(c) The distance between incentre and orthocentre is
\(\sqrt{2 r^{2}-4 R^{2} \cos A \cos B \cos C}\) - Perimeter (P) and area (A) of a regular polygon of n sides inscribed in a circle of radius r are given by
\(p=2 n r \sin \frac{\pi}{n} \text { and } A=\frac{1}{2} n r^{2} \sin \frac{2 \pi}{n}\) Perimeter and area ofa regular polygon ofn sides circumscribed about a given circle of radius r is given by
\(\mathrm{P}=2 \mathrm{nr} \tan \frac{\pi}{\mathrm{n}} \quad \text { and } \quad \mathrm{A}=\mathrm{nr}^{2} \tan \frac{\pi}{\mathrm{n}}\) - In many kinds of trignometric calculation, as in the solution of triangles, we often require the logarithms oftrignometrical ratios . To avoid the trouble and inconvenience ofprinting the proper sign to the logarithms ofthe trignometric functions, the logarithms as tabulated are not the true logarithms, but the true logarithms increased by 10 . The symbol L is used to denote these “tabular logarithms” . Thus: L sin 15º 25′ = 10 + log10 sin 15º 25′ and L tan 48º 23′ = 10 + log10 tan 48º 23′.