Find the value of x in the triangle. A. 7°B. 293°C. 113°D. 67°

Mathematics

2 answers:

andrezito [222]3 years ago

6 0

180-(60+53)
180-113
x=67

olasank [31]3 years ago

4 0

Hello there!

We know there are 150 degrees in a triangle

we already know 2 of them

60 and 53

We must add these together and subtract that from 180 to find our last angle

60 + 53 = 113
180 - 113 = 67

The final angle is 67 degrees.

I hope I helped!

~ Fire

You might be interested in

Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the cir

Bas_tet [7]

To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.

If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.

For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2

A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]

B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]

C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]