3.8.7 Pengamiran, SPM Praktis (Kertas 2)

Posted on May 13, 2020 by Myhometuition

Soalan 8:
Rajah di bawah menunjukkan lengkung

$y = \frac{4}{x^{2}}$ dan garis lurus y = mx + c. Garis lurus y = mx + c ialah tangen kepada lengkung pada (2, 1).

(a) Cari nilai m dan nilai c.

(b) Hitung luas kawasan berlorek.

(c) Diberi bahawa isi padu kisaran apabila rantau yang dibatasi oleh lengkung, paksi-x, garis lurus x = 2 dan x = h diputarkan melalui 360^o pada paksi-x ialah

$\frac{38 π}{81} {unit}^{3} .$
Cari nilai h, dengan keadaan h > 2.

Penyelesaian:
(a)

$\begin{array}{l} y = \frac{4}{x^{2}} = 4 x^{- 2} \\ \frac{d y}{d x} = - 8 x^{- 3} \\ = - \frac{8}{x^{3}} \\ At x = 2, \\ \frac{d y}{d x} = - \frac{8}{2^{3}} \\ = - 1 \\ Persamaan tangen: \\ y - y_{1} = m (x - x_{1}) \\ y - 1 = - 1 (x - 2) \\ y = - x + 2 + 1 \\ y = - x + 3 \\ m = - 1, c = 3 \end{array}$

(b)

$\begin{array}{l} Pada paksi- x, y = 0 \\ Dari garis lurus y = - x + 3, x = 3 \\ Luas kawasan berlorek \\ = Luas bawah lengkung - Luas segi tiga \\ = \int_{2}^{4} y d x - \frac{1}{2} \times 1 \times 1 \\ = \int_{2}^{4} (4 x^{- 2}) d x - \frac{1}{2} \\ = {[\frac{4 x^{- 1}}{- 1}]}_{2}^{4} - \frac{1}{2} \\ = {[- \frac{4}{x}]}_{2}^{4} - \frac{1}{2} \\ = [- \frac{4}{4} - (- \frac{4}{2})] - \frac{1}{2} \\ = \frac{1}{2} {unit}^{2} \end{array}$

(c)

$\begin{array}{l} Isipadu kisaran = \frac{38 π}{81} \\ π \int_{2}^{h} y^{2} d x = \frac{38 π}{81} \\ \int_{2}^{h} {(\frac{4}{x^{2}})}^{2} d x = \frac{38}{81} \\ \int_{2}^{h} (\frac{16}{x^{4}}) d x = \frac{38}{81} \\ \int_{2}^{h} (16 x^{- 4}) d x = \frac{38}{81} \\ {[\frac{16 x^{- 3}}{- 3}]}_{2}^{h} = \frac{38}{81} \\ {[- \frac{16}{3 x^{3}}]}_{2}^{h} = \frac{38}{81} \\ - \frac{16}{3 h^{3}} - (- \frac{16}{3 {(2)}^{3}}) = \frac{38}{81} \\ \frac{16}{3 h^{3}} = \frac{16}{24} - \frac{38}{81} \\ \frac{16}{3 h^{3}} = \frac{16}{81} \\ 3 h^{3} = 81 \\ h^{3} = 27 \\ h = 3 \end{array}$

3.8.6 Pengamiran, SPM Praktis (Kertas 2)

Posted on May 13, 2020 by Myhometuition

Soalan 7:
Rajah di bawah menunjukkan suatu lengkung

$y = \frac{1}{4} x^{2} + 3$ yang menyilang suatu garis lurus y = x + 6 pada titik A.

(a) Cari koordinat A.
(b) ) hitung
(i) luas rantau berlorek M,
(ii) isipadu kisaran, dalam sebutan π, apabila rantau berlorek N diputarkan melalui 360^o pada paksi-y.

Penyelesaian:
(a)

$\begin{array}{l} y = \frac{1}{4} x^{2} + 3.......... (1) \\ y = x + 6.......... (2) \\ Gantikan (2) ke dalam (1), \\ x + 6 = \frac{1}{4} x^{2} + 3 \\ 4 x + 24 = x^{2} + 12 \\ x^{2} - 4 x - 12 = 0 \\ (x + 2) (x - 6) = 0 \\ x = - 2 or x = 6 (ditolak) \\ Apabila x = - 2 \\ y = - 2 + 6 = 4 \\ Oleh itu, A = (- 2, 4) . \end{array}$

(b)(i)

$\begin{array}{l} Pada paksi- x, y = 0 \\ Dari y = x + 6, x = - 6 \\ Luas kawasan berlorek M \\ = Luas segi tiga + Luas di bawah lengkung \\ = \frac{1}{2} \times (6 - 2) \times 4 + \int_{- 2}^{0} y d x \\ = 8 + \int_{- 2}^{0} (\frac{1}{4} x^{2} + 3) d x \\ = 8 + {[\frac{x^{3}}{4 (3)} + 3 x]}_{- 2}^{0} \\ = 8 + [0 - (\frac{{(- 2)}^{3}}{12} + 3 (- 2))] \\ = 8 + [0 - (- \frac{8}{12} - 6)] \\ = 8 + [0 - (- \frac{20}{3})] \\ = 14 \frac{2}{3} {unit}^{2} \end{array}$

(b)(ii)

$\begin{array}{l} pada paksi- y, x = 0, \\ y = \frac{1}{4} (0) + 3 \\ y = 3 \\ y = \frac{1}{4} x^{2} + 3 \\ 4 y = x^{2} + 12 \\ x^{2} = 4 y - 12 \\ Isipadu N \\ = π \int_{3}^{4} x^{2} d y \\ = π \int_{3}^{4} (4 y - 12) d y \\ = π \int_{3}^{4} (2 y^{2} - 12 y) d y \\ = π {[(2 y^{2} - 12 y)]}_{3}^{4} \\ = π [(2 {(4)}^{2} - 12 (4)) - (2 {(3)}^{2} - 12 (3))] \\ = π (- 16 + 18) \\ = 2 π {unit}^{3} \end{array}$

3.7.5 Pengamiran, SPM Praktis (Kertas 1)

Posted on May 13, 2020 by Myhometuition

Soalan 11:

$\begin{array}{l} Diberi = \int_{2}^{5} g (x) d x = - 2 . Cari \\ (a) nilai bagi \int_{5}^{2} g (x) d x, \\ (b) nilai bagi m jika \int_{2}^{5} [g (x) + m (x)] d x = 19 \end{array}$

Penyelesaian:

$\begin{array}{l} (a) \int_{5}^{2} g (x) d x = - \int_{2}^{5} g (x) d x \\ = - (- 2) \\ = 2 \end{array}$

$\begin{array}{l} (b) \int_{2}^{5} [g (x) + m (x)] d x = 19 \\ \int_{2}^{5} g (x) d x + m \int_{2}^{5} x d x = 19 \\ - 2 + m {[\frac{x^{2}}{2}]}_{2}^{5} = 19 \\ \frac{m}{2} {[x^{2}]}_{2}^{5} = 21 \\ \frac{m}{2} [25 - 4] = 21 \\ 21 m = 42 \\ m = 2 \end{array}$

Soalan 12:

$\begin{array}{l} a) Cari nilai bagi \int_{- 1}^{1} {(3 x + 1)}^{3} d x . \\ (b) Nilaikan \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x . \end{array}$

Penyelesaian:

$\begin{array}{l} a) {\int_{- 1}^{1} {(3 x + 1)}^{3} d x = [\frac{{(3 x + 1)}^{4}}{4 (3)}]}_{- 1}^{1} \\ = {[\frac{{(3 x + 1)}^{4}}{12}]}_{- 1}^{1} \\ = \frac{1}{12} [4^{4} - {(- 2)}^{4}] \\ = \frac{1}{12} (256 - 16) \\ = 20 \end{array}$

$\begin{array}{l} (b) \int_{3}^{4} \frac{1}{\sqrt{2 x - 4}} d x = \int_{3}^{4} \frac{1}{{(2 x - 4)}^{\frac{1}{2}}} d x \\ = \int_{3}^{4} {(2 x - 4)}^{- \frac{1}{2}} d x \\ = {[\frac{{(2 x - 4)}^{- \frac{1}{2} + 1}}{\frac{1}{2} (2)}]}_{3}^{4} \\ = {[\sqrt{2 x - 4}]}_{3}^{4} \\ = [\sqrt{2 (4) - 4} - \sqrt{2 (3) - 4}] \\ = 2 - \sqrt{2} \end{array}$

Soalan 13:

$\begin{array}{l} Diberi y = \frac{x^{2}}{2 x - 1}, tunjukkan \\ \frac{d y}{d x} = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} . Seterusnya, nilaikan \int_{- 2}^{2} \frac{x (x - 1)}{4 {(2 x - 1)}^{2}} d x . \end{array}$

Penyelesaian:

$\begin{array}{l} y = \frac{x^{2}}{2 x - 1} \\ \frac{d y}{d x} = \frac{(2 x - 1) (2 x) - x (2)}{{(2 x - 1)}^{2}} \\ = \frac{4 x^{2} - 2 x - 2 x^{2}}{{(2 x - 1)}^{2}} \\ = \frac{2 x^{2} - 2 x}{{(2 x - 1)}^{2}} \\ = \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} (tertunjuk) \\ \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{8} \int_{- 2}^{2} \frac{2 x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} {[\frac{x^{2}}{2 x - 1}]}_{- 2}^{2} \\ \frac{1}{4} \int_{- 2}^{2} \frac{x (x - 1)}{{(2 x - 1)}^{2}} d x = \frac{1}{8} [(\frac{2^{2}}{2 (2) - 1}) - (\frac{{(- 2)}^{2}}{2 (- 2) - 1})] \\ = \frac{1}{8} [(\frac{4}{3}) - (\frac{4}{- 5})] \\ = \frac{1}{8} (\frac{32}{15}) \\ = \frac{4}{15} \end{array}$

3.7.4 Pengamiran, SPM Praktis (Kertas 1)

Posted on May 13, 2020 by Myhometuition

Soalan 8:

$\begin{array}{l} Diberi \int_{- 2}^{3} g (x) d x = 4, dan \int_{- 2}^{3} h (x) d x = 9, cari nilai bagi \\ (a) \int_{- 2}^{3} 5 g (x) d x, \\ (b) m jika \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \end{array}$

Penyelesaian:
(a)

$\begin{array}{l} \int_{- 2}^{3} 5 g (x) d x = 5 \int_{- 2}^{3} g (x) d x \\ = 5 \times 4 \\ = 20 \end{array}$

(b)

$\begin{array}{l} \int_{- 2}^{3} [g (x) + 3 h (x) + 4 m] d x = 12 \\ \int_{- 2}^{3} g (x) d x + 3 \int_{- 2}^{3} h (x) d x + \int_{- 2}^{3} 4 m d x = 12 \\ 4 + 3 (9) + 4 m {[x]}_{- 2}^{3} = 12 \\ 4 m [3 - (- 2)] = - 19 \\ 20 m = - 19 \\ m = - \frac{19}{20} \end{array}$

Soalan 9:

$Diberi y = \frac{5 x}{x^{2} + 1} dan \frac{d y}{d x} = g (x), cari nilai bagi \int_{0}^{3} 2 g (x) d x .$

Penyelesaian:

$\begin{array}{l} Memandangkan \frac{d y}{d x} = g (x), maka y = \int g (x) d x \\ \int_{0}^{3} 2 g (x) d x = 2 \int_{0}^{3} g (x) d x \\ = 2 {[y]}_{0}^{3} \\ = 2 {[\frac{5 x}{x^{2} + 1}]}_{0}^{3} \\ = 2 [\frac{5 (3)}{3^{2} + 1} - 0] \\ = 2 (\frac{15}{10}) \\ = 3 \end{array}$

Soalan 10:

$Cari \int_{5}^{k} (x + 1) d x, dalam sebutan k .$

Penyelesaian:

$\begin{array}{l} {\int_{5}^{k} (x + 1) d x = [\frac{x^{2}}{2} + x]}_{5}^{k} \\ = (\frac{k^{2}}{2} + k) - (\frac{5^{2}}{2} + 5) \\ = \frac{k^{2} + 2 k}{2} - \frac{35}{2} \\ = \frac{k^{2} + 2 k - 35}{2} \end{array}$

3.7.3 Pengamiran, SPM Praktis (Kertas 1)

Posted on May 13, 2020 by Myhometuition

Soalan 5:

$\begin{array}{l} Diberi \int (6 x^{2} + 1) d x = m x^{3} + x + c, \\ dengan keadaan m dan c ialah pemalar, cari \\ (a) nilai m . \\ (b) nilai c jika \int (6 x^{2} + 1) d x = 13 apabila x = 1. \end{array}$

Penyelesaian:
(a)

$\begin{array}{l} \int (6 x^{2} + 1) d x = m x^{3} + x + c \\ \frac{6 x^{3}}{3} + x + c = m x^{3} + x + c \\ 2 x^{3} + x + c = m x^{3} + x + c \\ Banding kedua-dua belah, \\ Maka, m = 2 \end{array}$

(b)

$\begin{array}{l} \int (6 x^{2} + 1) d x = 13 apabila x = 1. \\ 2 {(1)}^{3} + 1 + c = 13 \\ 3 + c = 13 \\ c = 10 \end{array}$

Soalan 6:

$\begin{array}{l} Diberi bahawa \int_{5}^{k} g (x) d x = 6, dan \int_{5}^{k} [g (x) + 2] d x = 14, \\ cari nilai k . \end{array}$

Penyelesaian:

$\begin{array}{l} \int_{5}^{k} [g (x) + 2] d x = 14 \\ \int_{5}^{k} g (x) d x + \int_{5}^{k} 2 d x = 14 \\ 6 + {[2 x]}_{5}^{k} = 14 \\ 2 (k - 5) = 8 \\ k - 5 = 4 \\ k = 9 \end{array}$

Soalan 7:

$Diberi \int_{k}^{2} (4 x + 7) d x = 28, hitung nilai yang mungkin bagi k .$

Penyelesaian:

$\begin{array}{l} \int_{k}^{2} (4 x + 7) d x = 28 \\ {[2 x^{2} + 7 x]}_{k}^{2} = 28 \\ 8 + 14 - (2 k^{2} + 7 k) = 28 \\ 22 - 2 k^{2} - 7 k = 28 \\ 2 k^{2} + 7 k + 6 = 0 \\ (2 k + 3) (k + 2) = 0 \\ k = - \frac{3}{2} atau k = - 2 \end{array}$

1.4.4 Janjang, SPM Praktis (Soalan Panjang)

Posted on May 13, 2020 by Myhometuition

Soalan 4:
Sebutan ketiga dan keenam suatu janjang geometri masing-masing ialah 24 dan

$7 \frac{1}{9}$ . Cari
(a) Sebutan pertama dan nisbah sepunya,
(b) Hasil tambah lima sebutan pertama,
(c) Hasil tambah n sebutan pertama dengan n yang cukup besar hingga rⁿ ≈ 0.

Penyelesaian:
(a)

$\begin{array}{l} Diberi T_{3} = 24 \\ a r^{2} = 24 ........... (1) \\ Diberi T_{6} = 7 \frac{1}{9} \\ a r^{5} = \frac{64}{9} ........... (2) \\ \frac{(2)}{(1)} : \frac{a r^{5}}{a r^{2}} = \frac{\frac{64}{9}}{24} \\ r^{3} = \frac{8}{27} \\ r = \frac{2}{3} \end{array}$

$\begin{array}{l} Gantikan r = \frac{2}{3} ke dalam (1) \\ a {(\frac{2}{3})}^{2} = 24 \\ a (\frac{4}{9}) = 24 \\ a = 24 \times \frac{9}{4} \\ = 54 \\ Jadi, sebutan pertama ialah 54 dan \\ nisbah sepunya ialah \frac{2}{3} . \end{array}$

(b)

$\begin{array}{l} S_{5} = \frac{54 [1 - {(\frac{2}{3})}^{5}]}{1 - \frac{2}{3}} \\ = 54 \times \frac{211}{243} \times \frac{3}{1} \\ = 140 \frac{2}{3} \\ Jadi, hasil tambah lima sebutan pertama \\ ialah 140 \frac{2}{3} . \end{array}$

(c)

$\begin{array}{l} Apabila - 1 < r < 1 dan n menjadi \\ cukup besar sehingga r^{n} \approx 0, \\ maka S_{n} = \frac{a}{1 - r} \\ = \frac{54}{1 - \frac{2}{3}} \\ = 162 \end{array}$

Jadi, hasil tambah n sebutan pertama dengan n yang cukup besar sehingga rⁿ ≈ 0 ialah 162.

1.4.3 Janjang, SPM Praktis (Soalan Panjang)

Posted on May 13, 2020 by Myhometuition

Soalan 3:
Suatu janjang aritmetik mempunyai 16 sebutan. Hasil tambah 16 sebutan itu ialah 188, manakala hasil tambah bagi sebutan-sebutan genap ialah 96. Cari
(a) sebutan pertama dan beza sepunya,
(b) sebutan terakhir.

Penyelesaian:
(a)
Katakan sebutan pertama = a
Beza sepunya = d

$\begin{array}{l} Diberi S_{16} = 188 \\ Maka, \frac{16}{2} [2 a + 15 d] = 188 \\ 8 [2 a + 15 d] = 188 \\ 2 a + 15 d = \frac{188}{8} \\ 2 a + 15 d = 23.5 - - - (1) \end{array}$

Diberi hasil tambah sebutan-sebutan genap = 96

$\begin{array}{l} T_{2} + T_{4} + T_{6} + ..... + T_{16} = 96 \\ (a + d) + (a + 3 d) + (a + 5 d) + ..... + (a + 15 d) = 96 \\ \frac{8}{2} [(a + d) + (a + 15 d)] = 96 \\ 4 [2 a + 16 d] = 96 \\ 2 a + 16 d = 24 - - - (2) \end{array}$

(2) – (1):
16d – 15d = 24 – 23.5
d = 0.5

Gantikan d = 0.5 ke dalam (2):
2a + 16 (0.5) = 24
2a + 8 = 24
2a = 16
a = 8
Maka, sebutan pertama = 8 dan beza sepunya = 0.5.

(b)
Sebutan terakhir = T₁₆
= 8 + 15 (0.5)
= 8 +7.5
= 15.5

10.4.8 Penyelesaian Segitiga, SPM Praktis (Kertas 2)

Posted on May 13, 2020 by Myhometuition

Soalan 8:
Rajah menunjukkan sisi empat kitaran PQRS.

(a) Hitung
(i) panjang, dalam cm, bagi PR,
(ii) ∠PRQ.
(b) Cari
(i) luas, dalam cm², bagi ∆ PRS.
(ii) jarak terdekat, dalam cm, dari titik S ke PR.

Penyelesaian:
(a)(i)

$\begin{array}{l} P R^{2} = 7^{2} + 8^{2} - 2 (7) (8) k o s 80^{o} \\ P R^{2} = 113 - 19.4486 \\ P R = \sqrt{93.5514} \\ P R = 9.6722 cm \end{array}$

(a)(ii)

$\begin{array}{l} Dalam sisi empat kitaran \\ ∠ P Q R + ∠ P S R = 180 \\ ∠ P Q R + 80 = 180 \\ ∠ P Q R = 100^{o} \\ \frac{\sin ∠ Q P R}{3} = \frac{\sin 100}{9.6722} \\ \sin ∠ Q P R = 0.3055 \\ ∠ Q P R = 17^{o} 47' \\ ∠ P R Q = 180^{o} - 100^{o} - 17^{o} 47' \\ = 62^{o} 13' \end{array}$

(b)(i)

$\begin{array}{l} Luas △ P R S \\ = \frac{1}{2} \times 7 \times 8 \sin 80^{o} \\ = 27.5746 {cm}^{2} \end{array}$

(b)(ii)

$\begin{array}{l} Luas △ P R S = 27.5746 \\ \frac{1}{2} \times 9.6722 \times h = 27.5746 \\ h = \frac{27.5746 \times 2}{9.6722} \\ = 5.7018 cm \\ Jarak terdekat = 5.7018 cm \end{array}$

10.4.7 Penyelesaian Segitiga, SPM Praktis (Kertas 2)

Posted on May 13, 2020 by Myhometuition

Soalan 7:
Rajah di bawah menunjukkan sebuah sisi empat ABCD dengan keadaan sisi AB dan sisi CD adalah selari. ∠BAC ialah sudut cakah.

Diberi bahawa AB = 14 cm, BC = 27 cm, ∠ACB = 30^o dan AB : DC = 7 : 3.
Hitung
(a) ∠BAC.
(b) panjang, dalam cm, bagi pepenjuru BD.
(c) luas, dalam cm², bagi sisi empat ABCD.

Penyelesaian:
(a)

$\begin{array}{l} \frac{\sin ∠ B A C}{27} = \frac{\sin 30^{o}}{14} \\ \sin ∠ B A C = \frac{\sin 30^{o}}{14} \times 27 \\ \sin ∠ B A C = 0.9643 \\ ∠ B A C = {74.64}^{o} \\ ∠ B A C (cakah) = 180^{o} - {74.64}^{o} \\ = {105.36}^{o} \end{array}$

(b)

$\begin{array}{l} A B selari dengan D C \\ ∠ B A C = ∠ A C D \\ ∠ B C D = {105.36}^{o} + 30^{o} \\ = {135.36}^{o} \\ \frac{D C}{A B} = \frac{3}{7} \\ D C = \frac{3}{7} \times 14 cm \\ = 6 cm \\ B D^{2} = 27^{2} + 6^{2} - 2 (27) (6) k o s ∠ B C D \\ B D^{2} = 765 - 324 kos {135.36}^{o} \\ B D^{2} = 995.54 \\ B D = 31.55 cm \end{array}$

(c)

$\begin{array}{l} ∠ A B C = 180^{o} - 30^{o} - {105.36}^{o} \\ = {44.64}^{o} \\ A C^{2} = 27^{2} + 14^{2} - 2 (27) (14) k o s ∠ A B C \\ A C^{2} = 925 - 756 k o s {44.64}^{o} \\ A C^{2} = 387.08 \\ A C = 19.67 cm \\ Luas △ A B C \\ = \frac{1}{2} (14) (27) \sin {44.64}^{o} \\ = 132.80 {cm}^{2} \\ Luas △ A C D \\ = \frac{1}{2} (19.67) (6) \sin {105.36}^{o} \\ = 56.90 {cm}^{2} \\ Luas sisi empat A B C D \\ = 132.80 + 56.90 \\ = 189.7 {cm}^{2} \end{array}$

10.4.6 Penyelesaian Segitiga, SPM Praktis (Kertas 2)

Posted on May 13, 2020 by Myhometuition

Soalan 6:
Rajah di bawah menunjukkan sebuah sisi empat PQRS.

(a) Cari
(i) panjang, dalam cm, bagi QS.
(ii) ∠QRS.
(iii) luas, dalam cm², bagi sisi empat PQRS.
(b)(i) Lakar sebuah segi tiga S’Q’R’ yang mempunyai bentuk berbeza daripada segi tiga SQR dengan keadaan S’R’ = SR, S’Q’ = SQ dan ∠S’Q’R’ = ∠SQR.
(ii) Seterusnya, nyatakan ∠S’R’Q’.

Penyelesaian:
(a)(i)

$\begin{array}{l} ∠ P = 180 - 76 - 34 = 70 \\ \frac{Q S}{\sin 70} = \frac{8}{\sin 34} \\ Q S = \frac{8 \times \sin 70}{\sin 34} \\ = 13.44 cm \end{array}$

(a)(ii)

$\begin{array}{l} {13.44}^{2} = 6^{2} + 9^{2} - 2 (6) (9) k o s ∠ Q R S \\ 108 k o s ∠ Q R S = 6^{2} + 9^{2} - {13.44}^{2} \\ k o s ∠ Q R S = \frac{6^{2} + 9^{2} - {13.44}^{2}}{108} \\ ∠ Q R S = k o s^{- 1} (- 0.5892) \\ = 126^{o} 6' \end{array}$

(a)(iii)

$\begin{array}{l} Luas P Q R S \\ = Luas P Q S + Luas Q R S \\ = (\frac{1}{2} \times 8 \times 13.44 \times \sin 76) + (\frac{1}{2} \times 6 \times 9 \times \sin 126^{o} 6') \\ = 52.16 + 21.82 \\ = 73.98 {cm}^{2} \end{array}$

(b)(i)

(b)(ii)

$\begin{array}{l} ∠ S' R' Q' = ∠ S' R R' \\ = 180 - 126^{o} 6' \\ = 53^{o} 54' \end{array}$