The perpendicular bisectors of two sides of a triangle meet at point that belongs to the third side. prove that this is a right triangle.

The point of intersection of the perpendicular bisectors of a triangle is the circumcenter, the center of the circle that contains the three triangle vertices.

If that center is on one side, that side must be a diameter of the circle. The diameter cuts the circle into two arcs, each of which measures 180°.

The third vertex of the triangle and its two legs form an inscribed angle that subtends an arc of 180°. The measure of that angle is half the measure of the arc, so the angle measures 90° and is a right angle.

A triangle with a right angle is a right triangle. QED

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)

step-by-step explanation: