Find question attached.​

a.x=50°

b.x=22°

Solution given:

a:

3x=2*75°[inscribed angle is half of central angle]

3x=150°

x=150°/3=50°

x=50°

b.

<BDC=34°+x[exterior angle is equal to the sum of two opposite interior angle of triangle]

again

<DCB=34°+x[base angle of isosceles triangle]

again

<ABC=90°[inscribed angle on a semi circle is 90°]

Now.

In triangle

ABC

<A+ <B+<C=180°[sum of interior angle of a triangle is 180°]

34°+90°+34°+x=180°

x=180°-90°-68°

x=90°-68°

x=22°

a) Solution

By using the inscribed angle is half of central angle,

→ 3x = 2 × 75

→ 3x = 150

→ x = 150/3

→ x = 50°

Thus, 50° is the value of x.

b) Solution

By using the exterior angle is equal to sum of two opposite interior angle of triangle,

→ <BDC = 34+x

→ <DCB = 34+x

(base angle of isosceles triangle)

→ <ABC = 90°

(inscribed angle on a semi circle is 90°)

Then in ∆ ABC,

By sum of interior angle of a triangle is 180°,

→ <A+<B+<C = 180°

→ 34+90+34+x=180°

→ x = 180°-90°-68°

→ x = 90°-68°

→ x = 22°

Thus, 22° is the value of x.

y = x + 95

y = 42 + 95

y=137 mm hg

a

step-by-step explanation: