Question

(challenger)(give answer with full explanation)

question 1.∠1 and ∠2 are complementary angles. ∠2 and ∠3 are supplementary angles. if ∠1 =45°, find the measure of ∠3.

question 2. one of the angle of an isosceles triangle is 100°. is this base angle or a vertical angle. give reasons. calculate the other two angles also.

Answer 1

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Qn(1).
If sum of two angles is 90° then these are complementary angles.
And,
If sum of two angles is 180°  then these are supplementary angles.
A/Q,
∠1 = 45° and ∠2 are complementary angles.
i.e,
    ∠1 + ∠2 = 90°
⇒ 45° + ∠2 = 90°
⇒  ∠2 = 90° - 45°
∴    ∠2 = 45°
and,
∠2 + ∠3 are supplementary angles.
i.e,
    ∠2 + ∠3 = 180°   
put the value of ∠2           [Calculated above]
⇒ 45° + ∠3 =180°
⇒ ∠3 = 180° - 45°
⇒ ∠3 = 135°

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Qn(2).

In a Isosceles triangle its two sides are equal and their corresponding angles are also equal.
Consider a triangle ΔABC,
Let ∠ABC & ∠ACB are base angles and ∠BAC is vertical angle.
∠ABC = ∠ACB    ≤ 0   
∠BAC ≤ 0  
We know that,
Sum of angles in a triangle is equal to 180°
⇒ ∠ABC + ∠ACB + ∠BAC = 180°
 ⇒  2.∠ABC  + ∠BAC = 180°      ----- ---------------(i)
Let the given angle i.e 100° is the base angle.
i.e ,∠ABC = ∠ACB = 100°
From eqn (i)
⇒  2×100 + ∠BAC =180°
⇒ 200 + ∠BAC= 180°
⇒ ∠BAC = 180° - 200.°
⇒ ∠BAC = -20°
Here ∠BAC  is  -ve, It is a contradiction to our assumption.
Hence given angle = 100° is not the base angle, it must be the vertical angle
i.e, 
∠ABC ≠ 100° & ∠ACB ≠ 100
∠BAC = 100°    [ vertical angle]
From Eqn (i)
2.∠ABC +  100° = 180 °  -----------------------(i)
⇒ 2∠ABC  = 180° -100°
⇒  2.∠ABC = 80°
⇒  ∠ABC =  80/2 = 40°
∴ ∠ABC =  40° & ∠ACB =  40°
Therefore vertical angle is 100° and base angle are 40° & 40°.

Answer 2

Step-by-step explanation:

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