Question

can someone plz solve in copy and send me pic?​​

Answer 1

Answer:

1. Angle ABD = Angle ACD.

Angle on same segment...

Angle ACD = 65

In ∆ AOC,

Angle AOC + Angle ACO + Angle CAO= 180. Angle of sum property,

90+65+ Angle CAO = 180

Angle CAO = 25

so angle CAB = 25

_________________________

2.

Given:

• PQ || ST, ∠PQR = 110° and ∠RST = 130°

To Find:

•  ∠QRS

Lines which are parallel same line are parallel to each other.

When two parallel lines are cut by a transversal, cco interior angles formed are supplementary.

Construction: Draw a line AB parallel to ST through point R. Since AB || ST and we know that PQ || ST. In the attacchment

Thus, AB || PQ.

Solution:

Let ∠SRQ = x, ∠SRB = y and ∠QRA = z

Therefore, the co-interior angles formed are supplementary.

∠RST + ∠SRB = 180°

130° + y = 180°

y = 180° - 130° = 50°

Thus, ∠SRB = y = 50°

∠PQR + ∠QRA = 180°

110° + z = 180°

z = 180° - 110° = 70°

Thus, ∠QRA = z = 70°

AB is a line, RQ and RS are rays on AB. Hence,

∠QRA + ∠QRS + ∠SRB = 180°

70° + x + 50° = 180°

120° + x = 180°

x = 180° - 120° = 60°

Thus, ∠QRS = x = 60°.

___________________________

3.

Given:

• Vertex angle is 45°

To find:

• Base angles of triangle?

Solution:

Now, we know that sum of the angles in a triangle is 180 degree,

$\sf{ \mapsto \: 2x + x + x = 180 \degree}$

$\sf{ \mapsto \: 4x = 180 \degree}$

$\sf{ \mapsto \: x = \frac{180}{4} }$

$\sf \pink{ \mapsto \: x = 45 \degree}$

Now the other angle is 2x

which is:-

$\sf \red{ \longmapsto \: 45 \times 2 = 90 \degree}$