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

∠B =
`55.00797173^(\circ)`

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=(perpendicular)/(base)`

Perpendicular = b =10

Base=a=7

Putting value in formula:

`tanB=(10)/(7)`

`tanB=1.42857`

`B=tan^(-1)1.42857`

`B=55.00797173^(\circ)`

Thus ∠B =
`55.00797173^(\circ)`