Question

help kn both........​

Answer 1

Step-by-step explanation:

hope this helps thank you

Answer 2

10.Given:

ΔPQR

Bisectors = ∠Q and ∠R

To Find:

∠QMR = 90 + ∠P/2,

Solution:

Bisectors of ∠Q and ∠R meet at point M.

Let ∠PQR =∠Q, ∠QRP =∠R and ∠QPR = 2P

In ΔPQR

∠P +∠Q+∠R = 180°  (sum of angles of a triangle)

= ( ∠Q +∠R = 180 − P ) --- eq 1

In ΔMQR

∠QMR + ∠Q/2 + ∠R/2 = 180°  (sum of angles of a triangle)

= (∠Q/2 + ∠R/2 = 180 - ∠QMR)

Thus, from equation 1 -

(180−∠P) /2  = 180 −∠QMR

90 - ∠P/2 = 180 −∠QMR

∠QMR = 90 + ∠P/2

Answer: ∠QMR = 90 + ∠P/2, hence proved

11.x=28, y=114, z=24

Step-by-step explanation:

in the given fig.,

CD is a straight line. so,

86+x+3z-6=180

x+3z=100 -----1

AB ia also a straight line,

so .,

x+86+2x+10=180

::x=28

substitute x=28 in equation 1,

we get z=24

in CD straight line,

2x+10+y=180

we already have x=28,

so, ::y=114