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secondary 3 | E Maths
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Candice lim
Candice Lim

secondary 3 chevron_right E Maths chevron_right Singapore

Hi, could you kindly advise me the working solution for this question please? It may be a simple question to you but this chapter is new to me so I am still struggling to learn :(
I can't solve it.
Have been struggling for the past 1hr.
Pls help and really thanks a lot!

Date Posted: 3 years ago
Views: 245
J
J
3 years ago
∠PTQ = ∠PSQ = 20°
(angles in the same segment, PQ)


∠PQT = 180° - ∠PTQ - ∠TPQ
= 180° - 20° - 100°
= 60° (angle sum of triangle PQT)
Candice lim
Candice Lim
3 years ago
Yes! Thanks a lot J :D
I didn't see it till you point it out to me.
haha...I think I know how to solve it now.

Thanks ;)
J
J
3 years ago
Welcome.


Alternative method :

∠TSQ + ∠TPQ = 180°

(angles in opposite segments are supplementary / opposite angles of a cyclic quadrilateral add up to 180°.

The quadrilateral here is TSQP )

So angle ∠TSQ

= 180° - ∠TPQ
= 180° - 100°
= 80°

∠TSP = ∠TSQ - ∠PSQ (adjacent angles)
= 80° - 20°
= 60°


∠PQT = ∠TSP = 60°

(angles in the same segment PT)
Candice lim
Candice Lim
3 years ago
Good morning J, thank you so much for your alternative solution :)

Have a nice day ahead.
J
J
3 years ago
Welcome

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Good evening Candice! Take this topic slowly, one step at a time. In this chapter, the difficulty lies in identifying the correct angles. Sometimes it does get difficult to spot a pattern inside the question.
Candice lim
Candice Lim
3 years ago
Good morning Mr Eric :)
Really appreciate your advice and once again thank you so much for your advised working solution.

Have a good day ahead.