please answer by solving
solv which you know
Answers
Step-by-step explanation:
Solutions:-
16)
In a Parallelogram ABCD , < A + <B = 180°
Since , The Adjacent angles in a Parallelogram are Supplementary.
17)
Given that
PQRS is a rhombus
The diagonals PR and QS intersects at O
We know that
In a rhombus the diagonals bisects at each other at 90°
So,
In ∆ POQ,
< POQ + <OPQ + <OQP = 180°
=> 90° + <OPQ + <OQP = 180°
=> <OPQ + <OQP = 180° -90°
=> <OPQ + <OQP = 90°
18)
ABCD is a Parallelogram.
AB is produced to E .
<CBE = 35°
We know that
<ABC + <CBE = 180°
Since , they are linear pair
=> <ABC + 35° = 180°
=> <ABC = 180°-35°
=> <ABC = 145°
We know that
In a Parallelogram, Opposite angles are equal.
=> The opposite angle of <ABC = <ADC
=> <ADC = 145°
Therefore, The measure of <ADC = 145°
19)
Given that
ABCD is a kite such that
AB = BC and AD = DC
Join A and C
Since AB = BC
=> <BAC = <BCA
Since , the angles opposite to equal sides are equal.
Let <BAC = <BCA = x°
and
Since AD = DC
=> <CAD = <ACD
Since , the angles opposite to equal sides are equal.
Let <CAD = <ACD = y°
Now
In ∆ ADC ,
< ADC + <CAD + <ACD = 180°
Since , the sum of the all angles in a triangle is 180°
=> 30° + y°+ y° = 180°
=> 30°+2y° = 180°
=> 2y° = 180°-30°
=> 2y° = 150°
=> y° = 150°/2
=> y° = 75°
Therefore, <CAD = <ACD = 75°
and
In ∆ ABC,
< ABC + <BAC + <BCA = 180°
Since , the sum of the all angles in a triangle is 180°
=> 115° + x° + x° = 180°
=> 115°+2x° = 180°
=> 2x° = 180°-115°
=> 2x° = 65°
=> x° = 65°/2
=> x° = 32.5°
Therefore, <BAC = <BCA = 32.5°
Now, < A = <BAC + <CAD
=> <A = 32.5° + 75°
=> <A = 107.5°
Therefore, <A = 107.5°
Answer:-
1) The measure of <ADC = 145°
2) The measure of <A = 107.5°
Used formulae:-
→ In a Parallelogram, Opposite angles are equal
→ The sum of two adjacent angles is 180° called Linear pair.
→ The sum of the all angles in a triangle is 180°
→ The angles opposite to equal sides are equal.
→ In a rhombus the diagonals bisects at each other at 90°
