Exercise Zone : Trigonometri

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Berikut ini adalah kumpulan soal mengenai Trigonometri. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSNBTHOTS

12

No. 1

Sebuah tangga disandarkan pada tembol rumah dengan membentuk sudut 60° terhadap tanah. Jarak antara ujung tangga dan permukaan tanah adalah $2\sqrt3$ m. Panjang tangga tersebut adalah .....

1. 4 m
2. 4,5 m

3. 5 m
4. 5,5 m

5. 6 m

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\sin 60\text{degree}& ={\displaystyle \frac{2\sqrt{3}}{x}}\\ {\displaystyle \frac{1}{2}}\sqrt{3}& ={\displaystyle \frac{2\sqrt{3}}{x}}\\ x& ={\displaystyle \frac{2\sqrt{3}}{{\displaystyle \frac{1}{2}}\sqrt{3}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$$

Jadi, panjang tangga tersebut adalah 4 m.
JAWAB: A

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No. 2

Jika $x-y=\dfrac12\pi$ maka tan x adalah....

1. $\dfrac{1+\tan y^2}{y}$
2. $-\dfrac{1-y^2}{\tan y}$
3. $\dfrac{\tan(1-y)}{(1+y)^2}$

4. $\dfrac{\tan y}{(1+y)^2}$
5. $-\dfrac1{\tan y}$

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ALTERNATIF PENYELESAIAN

$x=\dfrac12\pi+y$ $$\begin{array}{rl}\tan x& =\tan ({\displaystyle \frac{1}{2}}\pi +y)\\ & =-\cot y\\ & =-{\displaystyle \frac{1}{\tan y}}\end{array}$$

Jadi, $\tan x=-\dfrac1{\tan y}$.
JAWAB: E

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No. 3

Jika sin 13° = a, maka nilai cot 257° + csc 257 ° =

1. $\dfrac{a-1}{\sqrt{a^2-1}}$
2. $\dfrac{1-a}{\sqrt{a^2-1}}$
3. $\dfrac{a-1}{\sqrt{1-a^2}}$

4. $\dfrac{1-a}{\sqrt{1-a^2}}$
5. $\dfrac{-a-1}{\sqrt{1-a^2}}$

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\cot 257\text{degree}+\csc 257\text{degree}& =\cot (270\text{degree}-13\text{degree})+\csc (270\text{degree}-13\text{degree})\\ & =\tan 13\text{degree}-\sec 13\text{degree}\\ & ={\displaystyle \frac{a}{\sqrt{1-a^2}}}-{\displaystyle \frac{1}{\sqrt{1-a^2}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{a-1}{\sqrt{1-a^2}}}}}}}\end{array}$$

Jadi, $\cot257\degree+\csc257\degree=\dfrac{a-1}{\sqrt{1-a^2}}$.
JAWAB: C

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No. 4

Diketahui $\cos\alpha = \dfrac{a}{2b}$, dengan α sudut lancip dan b ≠ 0. Nilai dari tan α =

1. $\dfrac{2b}a$
2. $\dfrac{\sqrt{a^2-4b^2}}{2a}$
3. $\dfrac{\sqrt{4b^2-a^2}}{2a}$

4. $\dfrac{\sqrt{a^2-4b^2}}a$
5. $\dfrac{\sqrt{4b^2-a^2}}a$

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ALTERNATIF PENYELESAIAN

$\cos\alpha = \dfrac{a}{2b}=\dfrac{sa}{mi}$
sa = a, mi = 2b

$$\begin{array}{rl}de& =\sqrt{(2b{)}^2-a^2}\\ & =\sqrt{4b^2-a^2}\end{array}$$ $$\begin{array}{rl}\tan \alpha & ={\displaystyle \frac{de}{sa}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\sqrt{4b^2-a^2}}{a}}}}}}\end{array}$$

Jadi, $\tan\alpha=\dfrac{\sqrt{4b^2-a^2}}a$.
JAWAB: E

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No. 5

Jika θ sudut lancip dan $\cos\theta=\dfrac35$, maka nilai dari $\dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta}$ adalah

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ALTERNATIF PENYELESAIAN

$\sin\theta=\dfrac45$

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$\tan\theta=\dfrac43$ $$\begin{array}{rl}{\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}& ={\displaystyle \frac{\left({\displaystyle \frac{4}{5}}\right)\left({\displaystyle \frac{4}{3}}\right)-1}{2{\left({\displaystyle \frac{4}{3}}\right)}^2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{16}{15}}-1}{2\left({\displaystyle \frac{16}{9}}\right)}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{15}}}{{\displaystyle \frac{32}{9}}}}\\ & ={\displaystyle \frac{1}{15}}\cdot {\displaystyle \frac{9}{32}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{160}}}}}}\end{array}$$

Jadi, nilai dari $\dfrac{\sin\theta\tan\theta-1}{2\tan^2\theta}$ adalah $\dfrac3{160}$.

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No. 6

Nilai dari sin 135° adalah ....

1. $\dfrac13\sqrt2$
2. $\dfrac12\sqrt2$

3. $\dfrac14\sqrt2$
4. $\sqrt2$

5. $-\dfrac12\sqrt2$

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\sin 135°& =\sin (180°-45°)\\ & =\sin 45°\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{\displaystyle \frac{1}{2}}\sqrt{2}}}}}\end{array}$$

Jadi, nilai dari sin 135° adalah $\dfrac12\sqrt2$.
JAWAB: B

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No. 7

Diketahui $\sin \alpha=\dfrac45$, $0 \lt \alpha \lt \dfrac{\pi}2$ dan $\cos\beta =\dfrac{12}{13}$, $-\dfrac{\pi}2\lt\beta\lt0$. Tentukan nilai sin (α + β) dan sin (α − β).

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ALTERNATIF PENYELESAIAN

α kuadran I dan β kuadran IV

$\cos\alpha=\dfrac35$, $\sin\beta=-\dfrac5{13}$ $$\begin{array}{rl}\sin (\alpha +\beta )& =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & ={\displaystyle \frac{4}{5}}\cdot {\displaystyle \frac{12}{13}}+{\displaystyle \frac{3}{5}}\cdot (-{\displaystyle \frac{5}{13}})\\ & ={\displaystyle \frac{48}{65}}-{\displaystyle \frac{15}{65}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{33}{65}}}}}}\end{array}$$ $$\begin{array}{rl}\sin (\alpha -\beta )& =\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & ={\displaystyle \frac{4}{5}}\cdot {\displaystyle \frac{12}{13}}-{\displaystyle \frac{3}{5}}\cdot (-{\displaystyle \frac{5}{13}})\\ & ={\displaystyle \frac{48}{65}}+{\displaystyle \frac{15}{65}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{63}{65}}}}}}\end{array}$$

Jadi, $\sin (\alpha+\beta)=\dfrac{33}{65}$ dan $\sin (\alpha-\beta)=\dfrac{63}{65}$.

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No. 8

Segitiga ABC siku siku di C jika panjang AC = 20 cm dan besar sudut A = 60° tentukan panjang BC!

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\tan 60\text{degree}& ={\displaystyle \frac{BC}{AC}}\\ \sqrt{3}& ={\displaystyle \frac{BC}{20}}\\ BC& =\boxed{{\displaystyle \boxed{{\displaystyle 20\sqrt{3}}}}}\end{array}$$

Jadi, BC = $20\sqrt3$ cm.

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No. 9

Jika sudut θ di kuadran IV dan $\cos\theta=\dfrac1a$, maka sin θ = ....

1. $-\sqrt{a^2-1}$
2. $-\sqrt{1-a^2}$
3. $\dfrac{-1}{\sqrt{a^2-1}}$

4. $\dfrac{-\sqrt{a^2-1}}a$
5. $\dfrac{\sqrt{a^2-1}}a$

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ALTERNATIF PENYELESAIAN

θ di kuadran IV ⟶ a > 0

$\sin\theta=-\dfrac{\sqrt{a^2-1}}a$ ⟶ sinus bernilai negatif di kuadran IV.

Jadi, $\cot257\degree+\csc257\degree=\dfrac{a-1}{\sqrt{1-a^2}}$.
JAWAB: C

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No. 10

Bila 0° < a < 90° dan $\tan a=\dfrac5{\sqrt{11}}$, maka sin a = ....

1. $\dfrac56$
2. $\dfrac{25}{36}$

3. $\dfrac16\sqrt{11}$
4. $\dfrac5{36}$

5. $\dfrac1{36}\sqrt{11}$

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}x& =\sqrt{{\left(\sqrt{11}\right)}^2+5^2}\\ & =\sqrt{11+25}\\ & =\sqrt{36}\\ & =\boxed{{\displaystyle 6}}\end{array}$$ $$\begin{array}{rl}\sin a& ={\displaystyle \frac{5}{x}}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{\displaystyle \frac{5}{6}}}}}}\end{array}$$

Jadi, $\sin a=\dfrac56$.
JAWAB: A

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