Ask Singapore Homework?

Upload a photo of a Singapore homework and someone will email you the solution for free.



Question

secondary 3 | E Maths
One Answer Below

Anyone can contribute an answer, even non-tutors.

Answer This Question
Candy
Candy

secondary 3 chevron_right E Maths chevron_right Singapore

Can anyone help me with this qn?

Date Posted: 5 years ago
Views: 837
Muhd Ilman Fahmi
Muhd Ilman Fahmi
5 years ago
A) Take a look at the base rectangle ABCD. Two things to take note is that it's a rectangle, meaning the line BD cutting through the rectangle is a right angled triangle, since the corners of the rectangle are 90 degrees. Therefore using Pythagoras theorem, you can use the side lengths AB (29cm) and AD (22cm) to get the length DB, since ABD is a right angle triangle.

Second thing to take note is that it's a base, so the lines DQ and CP are perpendicular to the base. That makes triangle DBQ a right angled triangle. You have the length DB, and
DQ = CP= 15. So again, you can use Pythagoras to get BQ.

B) Looking at the triangle BPC, it's a right angled triangle with lengths 22 and 15. If you have learned trigonometry, you can now find the angle PBC using the inverse tangent of the two sides. Remember, it's opposite over adjacent, meaning it's the inverse tangent of 15/22.

C) Much like part B, you can use trigonometry to find the angle DBQ. Only this time you have all three lengths of the triangle DBQ. Use any of the trigonometry rules to find the answer.

Give it a try. Hope it helps!!
Candy
Candy
5 years ago
Thanks , i will try ! :)

See 1 Answer

done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
ThinkingYouths
Thinkingyouths's answer
15 answers (Tutor Details)
1st
Hello There,

The video working with answer is posted on our facebook page https://www.facebook.com/ThinkingYouths/videos/327614477843032/

If you need help with another question, upload your question as usual and type "need help from Thinking Youths" in the description box and we'll come to your aid.

Or visit www.ThinkingYouths.com to join us.