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Solutions of Triangle

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Estimated time: 14 minutes
Maharashtra State Board: Class 12

Formula: Coordinate Relations

  • x = r cosθ
  • y = r sinθ
  • \[\tan\theta=\frac{y}{x}\]
  • \[\mathbf{r}=\sqrt{x^2+y^2}\]
Maharashtra State Board: Class 12

Formula: Sine Rule

\[\mathrm{In~\Delta ABC,~\frac{a}{\sin A}=\frac{b}{sinB}=\frac{c}{sinC}}\]

Maharashtra State Board: Class 12

Formula: Cosine Rule

In ΔABC,

i.  \(\mathrm{cos}A=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2\mathrm{b}\mathrm{c}}\)

ii. \[\mathrm{cos}\mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2\mathrm{c}\mathrm{a}}\]

iii. \[\mathrm{cos}\mathrm{C}=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2\mathrm{ab}}\]

Maharashtra State Board: Class 12

Formula: Projection Rule

In ΔABC,

i. a = b cosC + c cosB
ii. b = c cosA + a cosC
iii. c = a cosB + b cosA

Maharashtra State Board: Class 12

Formula: Half Angle

In ΔABC, if a + b + c = 2s, then

1. \[\sin\frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s-b})(\mathrm{s-c})}{\mathrm{bc}}}\]

\[\sin\frac{\mathrm{B}}{2}=\sqrt{\frac{(\mathrm{s-c})(\mathrm{s-a})}{\mathrm{ca}}}\]

\[\sin\frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s-a})(\mathrm{s-b})}{\mathrm{ab}}}\]

2. \[\cos\frac{\mathrm{A}}{2}=\sqrt{\frac{\mathrm{s(s-a)}}{\mathrm{bc}}}\]

\[\cos\frac{\mathrm{B}}{2}=\sqrt{\frac{\mathrm{s(s-b)}}{\mathrm{ca}}}\]

\[\cos\frac{\mathrm{C}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}}\]

3. \[\tan\frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s-b})(\mathrm{s-c})}{\mathrm{s(s-a)}}}\]

\[\tan\frac{\mathrm{B}}{2}=\sqrt{\frac{(\mathrm{s-c})(\mathrm{s-a})}{\mathrm{s(s-b)}}}\]

\[\tan\frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s-a})(\mathrm{s-b})}{\mathrm{s(s-c)}}}\]

Maharashtra State Board: Class 12

Formula: Area of ΔABC

Area of ΔABC = \[\frac{1}{2}\mathrm{ab~sinC}\]

                         = \[=\frac{1}{2}\mathrm{bc~sinA}=\frac{1}{2}\mathrm{ac~sinB}\]

Heron’s Formula:

The area of ΔABC = \[\sqrt{\mathrm{s(s-a)(s-b)(s-c)}}\] 

where, 2s = a + b + c

Maharashtra State Board: Class 12

Formula: Napier’s Analogies

In ΔABC,

i. \[\tan\left(\frac{\mathrm{A-B}}{2}\right)=\left(\frac{\mathrm{a-b}}{\mathrm{a+b}}\right)\cot\frac{\mathrm{C}}{2}\]

ii. \[\tan\left(\frac{\mathrm{B-C}}{2}\right)=\left(\frac{\mathrm{b-c}}{\mathrm{b+c}}\right)\cot\frac{\mathrm{A}}{2}\]

iii. \[\tan\left(\frac{\mathrm{C-A}}{2}\right)=\left(\frac{\mathrm{c-a}}{\mathrm{c+a}}\right)\cot\frac{\mathrm{B}}{2}\]

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