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Answers
Step-by-step explanation:
Given -
AB = AC
∠ BAP = 60°
∠ CAP = 20°
∠ ACP = 10°
To Find -
Value of x Or ∠ APB = ?
Properties of Isosceles triangle-
- Isosecles Triangle consists of two equal sides, two equal angles, and the sum of internal angles of a triangle equal to 180°.
So,
This is an Isosceles triangle because here two sides are equal (AB = AC)
So,
In this triangle two angles are also equal.
Let ∠ABC = ∠ ACB = y
Then,
In ΔABC,
∠ BAC + ∠ ABC + ∠ ACB = 180°
= 80° + y + y = 180°
= 2y = 180° - 80°
= 2y = 100°
- = y = 50°
It means,
- ∠ABC = ∠ACB = 50°
∠ ACB = ∠ ACP + ∠ PCB
= 50° = 10° + < PCB
= ∠ PCB = 50° - 10°
- = ∠ PCB = 40°
Now,
In ΔACP,
∠ ACP + ∠ CAP + ∠ APC = 180°
= 10° + 20° + ∠ APC = 180°
= ∠ APC = 180° - 30°
- = ∠ APC = 150°
Since, point P is just a normal point inside ∆ABC,
Now, Either we have to Use Construction of Higher Level Geometry .
Or, we can use This by Trignometry
we will use Ceva Theorem in Trignometric Form :-
Let ∠ABP = z
( Since, ∠ABC = ∠ACB = 50°) .
By ceva theorem :-
→ sin60° * Sin10° * sin(50-z) = sinz * sin20° * sin40°
{ using sin20° = 2sin10*cos10°}
→ sin60° * sin(50-z) = [2sin40°*cos10°] * sinz
[ Now, use sinA*sin(60-A)sin(60+A) = (1/4)sin3A ]
You will get z = 20° .
_________________________
So ,
→ ∠x = 180°- (20+60) = 100° (Ans)..
Hence,
The value of x is 100°
Answer:
AB = AC
∠ BAP = 60°
∠ CAP = 20°
∠ ACP = 10°
To Find -
Value of x Or ∠ APB = ?
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Properties of Isosceles triangle-
Isosecles Triangle consists of two equal sides, two equal angles, and the sum of internal angles of a triangle equal to 180°.
So,
This is an Isosceles triangle because here two sides are equal (AB = AC)
So,
In this triangle two angles are also equal.
Let ∠ABC = ∠ ACB = y
Then,
In ΔABC,
∠ BAC + ∠ ABC + ∠ ACB = 180°
= 80° + y + y = 180°
= 2y = 180° - 80°
= 2y = 100°
= y = 50°
It means,
∠ABC = ∠ACB = 50°
∠ ACB = ∠ ACP + ∠ PCB
= 50° = 10° + < PCB
= ∠ PCB = 50° - 10°
= ∠ PCB = 40°
Now,
In ΔACP,
∠ ACP + ∠ CAP + ∠ APC = 180°
= 10° + 20° + ∠ APC = 180°
= ∠ APC = 180° - 30°
= ∠ APC = 150°
Since, point P is just a normal point inside ∆ABC,
Now, Either we have to Use Construction of Higher Level Geometry .
Or, we can use This by Trignometry
we will use Ceva Theorem in Trignometric Form :-
Let ∠ABP = z
( Since, ∠ABC = ∠ACB = 50°) .
By ceva theorem :-
→ sin60° * Sin10° * sin(50-z) = sinz * sin20° * sin40°
{ using sin20° = 2sin10*cos10°}
→ sin60° * sin(50-z) = [2sin40°*cos10°] * sinz
[ Now, use sinA*sin(60-A)sin(60+A) = (1/4)sin3A ]
You will get z = 20° .
_________________________
So ,
→ ∠x = 180°- (20+60) = 100° (Ans)..
Hence,
The value of x is 100°