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gulaghasi [49]
3 years ago
10

I got 2 questions.

Mathematics
2 answers:
alexandr402 [8]3 years ago
6 0

Answer:

1. In triangle ABC,

AC = a = 10, BC =b= 7 and ∠ C = 90°

By cosine law,

AB^2 = 7^2 + 10^2 - 2\times 7\times 10 cos 90^{\circ}

AB^2 = 49 + 100 - 0

AB^2 = 149

AB= \sqrt{149}

Now, by the law of sine,

\frac{sin B}{10} = \frac{sin90^{\circ}}{\sqrt{149} }

sin B = \frac{10}{\sqrt{149} }

\angle B = 55.0079798014\approx 55.008^{\circ}

2. In triangle ABC,

∠B = 30°,  AB=c=10 and ∠C = 90°

∠A = 180°-(30+90)°=60°

\frac{AC}{10}=sin30^{\circ}

⇒ AC = \frac{10}{2} = 5

By Pythagoras,

CB^2 = AB^2 - AC^2=10^2 - 5^2 = 100 - 25 = 75

⇒ CB = \sqrt{75} =8.66025403784\approx 8.66



Anit [1.1K]3 years ago
3 0
1) answer would be 12.2
2) answer would be 28.2888

hope this helps
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