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Penyelesaian Segitiga - Petua Sinus

Authored by ezDidik Edu

Mathematics

4th - 5th Grade

Used 9+ times

Penyelesaian Segitiga - Petua Sinus
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5 questions

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1.

MULTIPLE CHOICE QUESTION

10 mins • 1 pt

Media Image

Rajah menunjukkan sebuah segitiga ABC. Dengan menggunakan petua sinus, hitungkan nilai x.

12.034 cm

11.05 cm

16.98 cm

10.32 cm

Answer explanation

Media Image

a = x, A = 25, b = 15, B = 35

xsin25=15sin 35\frac{x}{\sin25}=\frac{15}{\sin\ 35}

x=15sin 35×sin 25x=\frac{15}{\sin\ 35}\times\sin\ 25


x=11.05 cmx=11.05\ cm

2.

MULTIPLE CHOICE QUESTION

10 mins • 1 pt

Media Image

Rajah menunjukkan segitiga ABC. Cari nilai x.

12.53 cm

13.53 cm

18.53 cm

11.43 cm

Answer explanation

Media Image

a = x, A = 125, b = 8, B = 35

xsin125=8sin35\frac{x}{\sin125}=\frac{8}{\sin35}

x=8sin35×sin125x=\frac{8}{\sin35}\times\sin125

x=11.43 cmx=11.43\ cm

3.

MULTIPLE CHOICE QUESTION

10 mins • 1 pt

Media Image

Rajah menunjukkan sebuah segitiga PQR dengan panjang QR ialah 700 m, PQR = 40°\angle PQR\ =\ 40^{\degree}  dan  PRQ=30°\angle PRQ=30^{\degree} . Cari panjang PQ.

412.54 m

366.14 m

372.46 m

445.90 m

Answer explanation

Media Image

a = PQ, A = 30, b = 700, B = 110

 PQsin30=700sin110\frac{PQ}{\sin30}=\frac{700}{\sin110}  

 PQ=700sin110×sin30PQ=\frac{700}{\sin110}\times\sin30 

 PQ=372.46 mPQ=372.46\ m  

4.

MULTIPLE CHOICE QUESTION

10 mins • 1 pt

Media Image

Cari nilai θ\theta  dengan menggunakan petua sinus.


 36.59°36.59\degree  

 45.98°45.98\degree  

 57.44°57.44\degree  

 23.69°23.69\degree  

Answer explanation

Media Image

A =  θ\theta  , a = 9.2, B = 100, b = 15.2

 sin θ9.2=sin10015.2\frac{\sin\ \theta}{9.2}=\frac{\sin100}{15.2}  
 sinθ=sin10015.2×9.2\sin\theta=\frac{\sin100}{15.2}\times9.2  
 θ=sin1(0.5961)\theta=\sin^{-1}\left(0.5961\right)  
 θ=36.59°\theta=36.59\degree  

5.

MULTIPLE CHOICE QUESTION

10 mins • 1 pt

Media Image

Cari nilai c.

25.16°25.16\degree

28.88°28.88\degree

30.25°30.25\degree

29.77°29.77\degree

Answer explanation

Media Image

A = c, a = 7.82, B = 110, b = 14.8

sin c7.82=sin11014.8\frac{\sin\ c}{7.82}=\frac{\sin110}{14.8}
sinc=sin11014.8×7.82\sin c=\frac{\sin110}{14.8}\times7.82
sinc=0.4965\sin c=0.4965
c=sin1(0.4965)c=\sin^{-1}\left(0.4965\right)
c=29.77°c=29.77\degree

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