Today’s question looks sooo simple. Just punch those keys on your state-of-the-art Casio scientific calculator and you will get it right … except that this is a Paper 1 question and your favourite Casio calculator should not be there!

So how now? Go ask your Grandma (Ah SOH) with the big foot (TOA CAH) for help lah!

In the diagram, a ladder BC is leaning against the wall AD. Given that, sin x = 5/13. Find the cosine of angle DCB.

The purpose of this question is to remind students that the standard trigo ratios (i.e. TOA CAH SOH -> Tan: Opp/Adj; Cos: Adj/Hypo; Sin: Opp/Hypo) applies only to an acute angle.

In order to solve for angle DCB which is an obtuse angle in the question, students need to remember a further set of equations i.e:

sin θ = sin (180^{o} - θ) cos θ = -cos (180^{o} - θ) tan θ = -tan (180^{o} - θ)

Hence exam questions usually require multiple steps like these. If everyday just simply TOA CAH SOH TOA CAH SOH, the examination board can close shop liao!

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

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## 2 Comments

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sin x = 5/13 => sin ABC = 5/13 => cos BCA = 5/13 => cos DCB = -5/13.

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Great stuff Nick!

The purpose of this question is to remind students that the standard trigo ratios (i.e. TOA CAH SOH -> Tan: Opp/Adj; Cos: Adj/Hypo; Sin: Opp/Hypo) applies only to an

acuteangle.In order to solve for angle DCB which is an

obtuseangle in the question, students need to remember a further set of equations i.e:sin θ = sin (180

^{o}- θ)cos θ = -cos (180

^{o}- θ)tan θ = -tan (180

^{o}- θ)Hence exam questions usually require multiple steps like these. If everyday just simply TOA CAH SOH TOA CAH SOH, the examination board can close shop

liao!So remember this well!