Question

1 solve triangle ABC which have
angleA=250.251,angleB=600.511anda=3.82.Findc.
angleA=250.251,angleB=600.511anda=3.82.Findc.



3-6cm
7.0cm
7.4cm
8.8cm




2 Solve triangle ABC which have
angleA=250.251,angleB=600.511anda=3.82.Findb.
angleA=250.251,angleB=600.511anda=3.82.Findb.



5.0cm
7.8cm
7.1cm
6.7cm

Answer 1

Answer on Question #61186 – Math – Trigonometry

Question

1. Solve triangle ABC which have angle A=250.251, angle B=600.511 and a=3.82. Find c.

3.6cm

7.0cm

7.4cm

8.8cm

2. Solve triangle ABC which have angle A=250.251, angle B=600.511 and a=3.82. Find b.

5.0cm

7.8cm

7.1cm

6.7cm

Solution

The values you've given for A and B are unrealistic, because the sum of angles of triangle should be $180{}^{\circ}$ (but only angle A=250.251°>180°).

But if there is a typo in the task and the real values of angles are A=25°25', B=60°51', then the problem can be solved by using law of sines:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$

So

1. Given $A = 25{}^{\circ}25'$, $B = 60{}^{\circ}51'$, $a = 3.82$. Then $C = 180{}^{\circ} - A - B = 180{}^{\circ} - 25{}^{\circ}25' - 60{}^{\circ}51' = 93{}^{\circ}44'$ and $c = \frac{a \cdot \sin C}{\sin A} = \frac{3.82 \cdot \sin(93{}^{\circ}44')}{\sin(25{}^{\circ}25')} = 8.88 \approx 8.9$.

2. Given $A = 25{}^{\circ}25'$, $B = 60{}^{\circ}51'$, $a = 3.82$. Then $C = 180{}^{\circ} - A - B = 180{}^{\circ} - 25{}^{\circ}25' - 60{}^{\circ}51' = 93{}^{\circ}44'$ and $b = \frac{a \cdot \sin B}{\sin A} = \frac{3.82 \cdot \sin(60{}^{\circ}51')}{\sin(25{}^{\circ}25')} = 7.77 \approx 7.8$.

**Answer:** 1. 8.9.

2. 7.8.

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