In △ABC, AB = AC = 15 cm, BC = 18 cm. Find (i) cos ∠ABC (ii) sin ∠ACB.

Question

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Draw AD perpendicular to BC. D is mid-point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side] ∴ BD = DC = 9 cm. In right-angled triangle ABD, ⇒ AB² = AD² + BD² ⇒ 15² = AD² + 9² ⇒ AD² = (15)² - (9)² ⇒ AD² = 225 - 81 ⇒ AD² = 144 ⇒ AD = = 12 cm. (i) cos ∠ABC = = . Hence, cos ∠ABC = . (ii) sin ∠ACB = = . Hence, sin ∠ACB = .

Mathematics

In △ABC, AB = AC = 15 cm, BC = 18 cm. Find
(i) cos ∠ABC
(ii) sin ∠ACB.

Trigonometrical Ratios

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Mathematics

In △ABC, AB = AC = 15 cm, BC = 18 cm. Find(i) cos ∠ABC(ii) sin ∠ACB.

Trigonometrical Ratios

Related Questions

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In △ABC, AB = AC = 15 cm, BC = 18 cm. Find
(i) cos ∠ABC
(ii) sin ∠ACB.

In ∆ABC, ∠A = 90°. If AB = 7 cm and BC - AC = 1 cm, find :
(i) sin C
(ii) tan B

In △PQR, ∠Q = 90°. If PQ = 40 cm and PR + QR = 50 cm, find :
(i) sin P
(ii) cos P
(iii) tan R

In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find
(i) sin C
(ii) tan B
(iii) tan C - cot B.

In the figure (2) given below, △ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin² θ + tan² θ.