Question

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

Answer 1

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

Answer 2

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 =$55.00797173^{\circ}$