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True [87]
3 years ago
10

In triangle ABC find b to the nearest degree given a=7 b=10 and c is a right triangle

Mathematics
2 answers:
GaryK [48]3 years ago
8 0
A^2+b^2=c^2

7^2+10^2=c^2
49+100=c
149=c
take the square root and the anser would be 12.2
so c=12.2
eimsori [14]3 years ago
4 0

Answer:

∠B =55.00797173^{\circ}

Step-by-step explanation:

Given : a=7 b=10 and C is a right triangle

solution :

refer the attached figure .

Since ΔACB is right angled triangle so we will use trigonometric ratios .

tan\theta=\frac{perpendicular}{base}

Perpendicular = b =10

Base=a=7

Putting value in formula:

tanB=\frac{10}{7}

tanB=1.42857

B=tan^{-1}1.42857

B=55.00797173^{\circ}

Thus ∠B =55.00797173^{\circ}

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