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NandishSS
KanishkM

IMO E

Now PQR is an isosceles triangle, < Q = 80

3 Cases
1) when <P = < R = 50
2) when <Q = <P = 80
3) when <Q = <R = 160, <P = 20

Sum = 150

Hey KanishkM,

3) when <Q = <R = 160, <P = 20 How is it possible? While Sum of angles of triangle is 180 ;)

hahahah NandishSS

I think i did some big time typo :shh:, let me correct it now
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This is an isosceles triangle, so two sides are equal and two angles must be of the same measure.

There are three cases:

1) angle Q is formed by two equal sides, so the other two angles (P and R) must be the same.

\(80 (angle Q) + angle P + angle R = 180\) (sum of all angles in a triangle)

\(angle P + angle R = 100\)

As mentioned before, angle P and angle R must be equal, hence P = 50 and R = 50.

2) angle Q is formed by the base and one of the sides, and the angle (let's say R) is formed by the base and the other side. It means, that Q=R, so \(P = 180-80-80 = 20\)

3) angle Q is formed by the base and one of the sides, and the angle (this time it's P) is formed by the base and the other side. It means, that Q=P, so P = 80

Sum of all possible values of angle \(P = 50 + 20 + 80 = 150\).

Answer E.
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In an isosceles triangle PQR, if ∠Q = 80 degrees, then what is the sum of all possible values of ∠P?

    A. 70
    B. 80
    C. 100
    D. 130
    E. 150

P can be equal to 80, if side PR is the same length as QR.

P can be 20 if side PQ is the same as PR

P can be 50 if side PQ and RQ are the same.

total is 150
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Solution


Given:
    • PQR is an isosceles triangle
    • ∠Q = 80 degrees

To find:
    • Sum of all possible values of ∠P

Approach and Working:
We know that in an isosceles triangle two angles are equal
    Case 1: If ∠Q = ∠P, then ∠P = 80 degrees
    Case 2: If ∠Q = ∠R, then ∠P = 180 – 80 – 80 = 20 degrees
    Case 3: If ∠P = ∠R, then ∠P = \(\frac{(180 – 80)}{2} = 50\) degrees

Therefore, sum of all possible values of ∠P = 80 + 50 + 20 = 150

Hence the correct answer is Option E.

Answer: E

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