How do you solve the triangle given m∠A = 70°, c = 26, a = 25?

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How do you solve the triangle given m∠A = 70°, c = 26, a = 25?

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Answer 1 · 2017-11-13T11:18:36

See below.

Explanation:

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From the information given, we can solve this using the Sine Rule:

The sine rule states that:

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $sinA/a=sinB/b=sinC/c$

Finding angle C. Since we know angle A and side a, we will use:

$\frac{sin A}{a} = \frac{sin C}{c}$ $sinA/a=sinC/c$

$\frac{sin (70^{o})}{25} = \frac{sin C}{26} \Rightarrow sin C = \frac{26 sin (70^{o})}{25} = 0.9778$ $sin(70^o)/25=sinC/26=>sinC=(26sin(70^o))/25=0.9778$

$C = {sin}^{- 1} (sin C) = {sin}^{- 1} (0.9778) = {77.76}^{o}$ $C=sin^-1(sinC)=sin^-1(0.9778)=color(blue)(77.76^o)$

Angle $B = 180^{o} - (70^{o} + {77.76}^{o}) = {32.24}^{o}$ $B = 180^o-(70^o +77.76^o)=color(blue)(32.24^o)$

Side b:

$\frac{sin (70^{o})}{25} = \frac{sin ({32.24}^{o})}{b} \Rightarrow b = \frac{25 sin ({32.24}^{o})}{sin (70^{o})} = 14.19$ $sin(70^o)/25=sin(32.24^o)/b=>b=(25sin(32.24^o))/sin(70^o)=color(blue)(14.19)$

So the solution is:

$A = 70^{o}$ $A=70^o$

$B = {32.24}^{o}$ $B=32.24^o$

$C = {77.76}^{o}$ $C=77.76^o$

$a = 25$ $a=25$

$b = 14.19$ $b=14.19$

$c = 26$ $c=26$

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How do you solve the triangle given m∠A = 70°, c = 26, a = 25?

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